General Stoichiometry Problems with Answers and Solutions – Chemistry Grade 11 SCH3U

Set 1 (Chemistry 20 – Gravimetric Stoichiometry)

Moles to Moles

1. Sulfur and Barium Oxide Reaction
   Given: Sulfur (S₈) reacts with barium oxide to produce barium sulfide and oxygen gas.
   • (a) How many moles of sulfur are needed if 2.00 mol of barium oxide are used?
   • (b) How many moles of barium sulfide are produced from 0.100 mol of sulfur?
2. Combustion of Methane
   Reaction: CH₄(g) + O₂(g) → CO₂(g) + H₂O(g)
   • How many moles of oxygen gas are needed to completely burn 4.00 mol of methane gas?
   • How many moles of water vapor are produced from 0.0400 mol of methane gas?
3. Sodium Metal and Phosphorus Reaction
   Given: Sodium metal (Na) reacts with phosphorus (P₄) to form a compound.
   • How many moles of phosphorus are needed if 0.600 mol of sodium metal are used?
   • How many moles of sodium phosphide can be produced from 0.25 mol of sodium metal?

Moles to Mass

4. Sodium Metal and Water Reaction
   Given: Sodium metal reacts rapidly with water in a single replacement reaction producing hydrogen gas and sodium hydroxide.
   Hint: The second product is NaOH(aq).
   • Question: What mass of sodium metal must react with water to produce 0.540 mol of hydrogen gas?
5. Slaked Lime and Carbon Dioxide Reaction
   Given: Slaked lime (Ca(OH)₂) is used for whitewashing or in mortar for bricklaying. It reacts with carbon dioxide in the air to produce calcium carbonate and water.
   • Question: What mass of calcium carbonate will be produced by the reaction of 0.962 mol of calcium hydroxide with carbon dioxide?

Mass to Moles

6. Sulfur to Sulfur Dioxide Conversion
   Given: In the first step of the production of sulfuric acid, sulfur is burned in air to produce sulfur dioxide.
   • Question: How many moles of oxygen are required to react with 160 g of sulfur to produce sulfur dioxide?
7. Production of Hydrogen Fluoride (HF)
   Given: Hydrogen fluoride (HF) may be produced in the lab and used to etch glass. HF fumes are irritating and corrosive.
   • Question: How many moles of hydrofluoric acid may be produced by the addition of 7.81 g of solid calcium fluoride to excess sulfuric acid? (Hint: This is a double replacement reaction.)

Mass to Mass

8. Refining Iron Ore (Fe₂O₃)
   Given: In the process of refining iron ore, the iron ore (Fe₂O₃) is reacted with coke (C), producing iron and carbon monoxide.
   • Question: What mass of coke is required for every 1.00 kg of iron ore refined?
9. Thermite Reaction
   Given: The thermite reaction is used for making crude welds. The mixture is composed of aluminum and iron(III) oxide.
   • Question: What mass of solid iron(III) oxide is required for every 40.5 g of solid aluminum? (Hint: This is a single replacement reaction.)

Thought about stoichiometry solutions for 24 seconds

Below are the detailed solutions to the second set of problems (Chemistry 20 – Gravimetric Stoichiometry). Each step explains the reasoning and calculations involved.

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SOLUTIONS Set 1: Moles to Moles

1. Sulfur (S₈) and Barium Oxide (BaO)

Given Reaction:
S₈ + 8 BaO → 8 BaS + 4 O₂

(a) How many moles of sulfur are needed if 2.00 mol of barium oxide are used?

Step-by-Step:

• The balanced equation shows a ratio of 1 mole S₈ to 8 moles BaO.
• Given: 2.00 mol BaO
• Using the ratio: moles S8=2.00 mol BaO×1 mol S88 mol BaO=0.250 mol S8\text{moles S}_8 = 2.00 \, \text{mol BaO} \times \frac{1 \, \text{mol S}_8}{8 \, \text{mol BaO}} = 0.250 \, \text{mol S}_8moles S8​=2.00mol BaO×8mol BaO1mol S8​​=0.250mol S8​

Answer (a): 0.250 mol S₈

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(b) How many moles of barium sulfide (BaS) are produced from 0.100 mol of S₈?

Step-by-Step:

• From the balanced equation: 1 mole S₈ produces 8 moles BaS.
• Given: 0.100 mol S₈
• Moles BaS: 0.100 mol S8×8 mol BaS1 mol S8=0.800 mol BaS0.100 \, \text{mol S}_8 \times \frac{8 \, \text{mol BaS}}{1 \, \text{mol S}_8} = 0.800 \, \text{mol BaS}0.100mol S8​×1mol S8​8mol BaS​=0.800mol BaS

Answer (b): 0.800 mol BaS

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2. Combustion of Methane

Reaction:
CH₄ + 2 O₂ → CO₂ + 2 H₂O

(a) How many moles of O₂ are needed to burn 4.00 mol CH₄?

• Ratio from the equation: 1 CH₄ : 2 O₂
• Given: 4.00 mol CH₄

4.00 mol CH4×2 mol O21 mol CH4=8.00 mol O24.00 \, \text{mol CH}_4 \times \frac{2 \, \text{mol O}_2}{1 \, \text{mol CH}_4} = 8.00 \, \text{mol O}_24.00mol CH4​×1mol CH4​2mol O2​​=8.00mol O2​

Answer (a): 8.00 mol O₂

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(b) How many moles of H₂O are produced from 0.0400 mol CH₄?

• Ratio: 1 CH₄ : 2 H₂O
• Given: 0.0400 mol CH₄

0.0400 mol CH4×2 mol H2O1 mol CH4=0.0800 mol H2O0.0400 \, \text{mol CH}_4 \times \frac{2 \, \text{mol H}_2\text{O}}{1 \, \text{mol CH}_4} = 0.0800 \, \text{mol H}_2\text{O}0.0400mol CH4​×1mol CH4​2mol H2​O​=0.0800mol H2​O

Answer (b): 0.0800 mol H₂O

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3. Sodium (Na) and Phosphorus (P₄)

Likely Reaction:
3 Na + P → Na₃P

(Phosphorus here can be considered as elemental P in a suitable form, often P₄, but the stoichiometry given suggests a simpler 3:1 ratio. The essential ratio is Na:P = 3:1.)

(a) How many moles of P are needed if 0.600 mol Na are used?

• Ratio: 3 Na : 1 P
• Given: 0.600 mol Na

0.600 mol Na×1 mol P3 mol Na=0.200 mol P0.600 \, \text{mol Na} \times \frac{1 \, \text{mol P}}{3 \, \text{mol Na}} = 0.200 \, \text{mol P}0.600mol Na×3mol Na1mol P​=0.200mol P

Answer (a): 0.200 mol P

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(b) How many moles of Na₃P can be produced from 0.25 mol Na?

• From equation: 3 Na → 1 Na₃P
• Given: 0.25 mol Na

0.25 mol Na×1 mol Na3P3 mol Na=0.0833… mol Na3P0.25 \, \text{mol Na} \times \frac{1 \, \text{mol Na}_3\text{P}}{3 \, \text{mol Na}} = 0.0833… \, \text{mol Na}_3\text{P}0.25mol Na×3mol Na1mol Na3​P​=0.0833…mol Na3​P

Answer (b): 0.0833 mol Na₃P (approximately)

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Moles to Mass

4. Sodium and Water Reaction

Reaction:
2 Na + 2 H₂O → 2 NaOH + H₂

For simplicity: 2 Na produce 1 H₂.

Given: We need 0.540 mol H₂.

• Ratio: 2 Na : 1 H₂
• For every 1 mol H₂, need 2 mol Na.

moles Na=0.540 mol H2×2 mol Na1 mol H2=1.08 mol Na\text{moles Na} = 0.540 \, \text{mol H}_2 \times \frac{2 \, \text{mol Na}}{1 \, \text{mol H}_2} = 1.08 \, \text{mol Na}moles Na=0.540mol H2​×1mol H2​2mol Na​=1.08mol Na

Mass of Na:

• M(Na)=22.99 g/mol

1.08 mol Na×22.99gmol=24.8 g Na1.08 \, \text{mol Na} \times 22.99 \frac{g}{\text{mol}} = 24.8 \, g \, \text{Na}1.08mol Na×22.99molg​=24.8gNa

Answer #4: 24.8 g Na

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5. Slaked Lime and CO₂ Reaction

Reaction:
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

• Ratio: 1:1 between Ca(OH)₂ and CaCO₃.

Given: 0.962 mol Ca(OH)₂moles CaCO3=0.962 mol Ca(OH)2×1 mol CaCO31 mol Ca(OH)2=0.962 mol CaCO3\text{moles CaCO}_3 = 0.962 \, \text{mol Ca(OH)}_2 \times \frac{1 \,\text{mol CaCO}_3}{1 \,\text{mol Ca(OH)}_2} = 0.962 \, \text{mol CaCO}_3moles CaCO3​=0.962mol Ca(OH)2​×1mol Ca(OH)2​1mol CaCO3​​=0.962mol CaCO3​

Mass of CaCO₃:

• M(CaCO₃) ≈ 100.09 g/mol

0.962 mol×100.09gmol≈96.3 g0.962 \, \text{mol} \times 100.09 \frac{g}{\text{mol}} \approx 96.3 \, g0.962mol×100.09molg​≈96.3g

Answer #5: 96.3 g CaCO₃

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Mass to Moles

6. Production of SO₂ from S

Reaction:
S + O₂ → SO₂

• Ratio: 1 S : 1 O₂

Given: 160 g S
M(S)=32.06 g/molmoles S=160 g32.06 g/mol≈4.99 mol S\text{moles S} = \frac{160 \, g}{32.06 \, g/mol} \approx 4.99 \, \text{mol S}moles S=32.06g/mol160g​≈4.99mol S

Therefore, moles O₂ needed = 4.99 mol (1:1 ratio)

Answer #6: 4.99 mol O₂

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7. Production of HF from CaF₂ and H₂SO₄

Reaction:
CaF₂ + H₂SO₄ → CaSO₄ + 2 HF

Ratio: 1 CaF₂ → 2 HF

Given: 7.81 g CaF₂
M(CaF₂)=40.08 (Ca) + 19×2 (F₂=38) = 78.08 g/molmoles CaF2=7.81 g78.08 g/mol≈0.1 mol CaF2\text{moles CaF}_2 = \frac{7.81 \, g}{78.08 \, g/mol} \approx 0.1 \, \text{mol CaF}_2moles CaF2​=78.08g/mol7.81g​≈0.1mol CaF2​

Moles HF =0.1 × 2 =0.2 mol

Answer #7: 0.20 mol HF

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Mass to Mass

8. Refining Iron Ore (Fe₂O₃) with Coke (C)

Reaction:
Fe₂O₃ + 3 C → 2 Fe + 3 CO

For every 1 mole Fe₂O₃, need 3 moles C.

Step-by-Step:

• Given mass: 1.00 kg Fe₂O₃ =1000 g Fe₂O₃
• M(Fe₂O₃)=Fe:55.85×2=111.7 + O₃:48 =159.7 g/mol approx.

moles Fe2O3=1000 g159.7 g/mol≈6.26 mol\text{moles Fe}_2\text{O}_3 = \frac{1000 \, g}{159.7 \, g/mol} \approx 6.26 \, \text{mol}moles Fe2​O3​=159.7g/mol1000g​≈6.26mol

Moles C needed =6.26 × 3=18.78 mol C
Mass C: 12.01 g/mol ×18.78 mol=225.7 g C

Answer #8: About 226 g of coke per 1.00 kg Fe₂O₃

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9. Thermite Reaction

Reaction:
2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe

Ratio: 2 Al :1 Fe₂O₃

Given: 40.5 g Al
M(Al)=26.98 g/molmoles Al=40.5 g26.98 g/mol≈1.50 mol Al\text{moles Al} = \frac{40.5 \, g}{26.98 \, g/mol} \approx 1.50 \, \text{mol Al}moles Al=26.98g/mol40.5g​≈1.50mol Al

From ratio: 2 Al per 1 Fe₂O₃
Fe₂O₃ needed = 1.50 mol Al × (1 Fe₂O₃ / 2 Al)=0.75 mol Fe₂O₃

M(Fe₂O₃)=159.7 g/mol (from previous calculation)
Mass Fe₂O₃=0.75 mol ×159.7 g/mol=119.8 g

Answer #9: About 120 g Fe₂O₃ per 40.5 g Al

SET 2

13. Foam-Produced Fire Extinguisher Reaction
Given: A foam-producing fire extinguisher should be used for fighting gasoline fires. The foam, Al(OH)₃, and the carbon dioxide produced both help to smother the fire.
Reaction: NaHCO₃(s) + Al₂(SO₄)₃(s) → Al(OH)₃(s) + CO₂(g) + Na₂SO₄(s)

• Question: Determine the mass of foam produced from 1.000 kg of baking soda.

14. Carbohydrate Combustion
Given: Carbohydrates (such as C₆H₁₂O₆) undergo combustion with oxygen to produce carbon dioxide and water. Living things use this reaction, called respiration, to produce energy.

• Question: Determine the mass of carbohydrate consumed for every 3.00 g of oxygen gas consumed during respiration.

15. Production of Copper by Zinc Strips
Given: Copper can be produced through a reaction between a zinc strip and a copper(II) sulfate solution.

• Question: What mass of copper should be produced? (Theoretical yield of copper is given as 15.42 g.)
• Given Data: Final mass of zinc = 14.15 g. (Hint: What mass of zinc reacted?)

16. Mass of Coal for Producing Gasoline
Given: When producing gasoline (assume C₈H₁₈) from coal (assume C) and hydrogen gas, some mass of coal will react with H₂ to form hydrocarbon.

• Question: What mass of coal is required for producing 1.00 kg of gasoline?

17. Producing Hydrogen Gas
Question: How many moles of chlorine gas are required to produce 1.000 kg of laundry bleach [Ca(OCl)₂·CaCl₂] from solid lime and calcium oxide?

18. Production of NaOH
Given: NaOH can be produced from the reaction of 148 g of slaked lime [Ca(OH)₂] with excess soda ash (Na₂CO₃).

• Question: What mass of NaOH can be produced?

19. Producing Poisonous Brown Gas from Copper
Given: When a copper strip is added to concentrated nitric acid, a brown poisonous gas is produced.
Reaction: Cu(s) + HNO₃(aq) → (something) + H₂O(l) + NO₂(g)

• Question: Determine the number of moles of gas produced when 1.27 g of copper reacts.

20. Saturn Rocket Fuel Reaction
Given: 5,000 kg of liquid hydrogen react with 5,000 kg of liquid oxygen in a Saturn rocket.

• Question: What mass of water is produced?

21. Gypsum Crystals (Plaster of Paris)
Given: Gypsum crystals “set” plaster of Paris. CaSO₄·2H₂O can be prepared from 29.04 g of plaster of Paris.
Reaction: CaSO₄·½H₂O + 3 H₂O → 2 CaSO₄·2H₂O

• Question: What mass of gypsum crystals (CaSO₄·2H₂O) can be prepared?

SET 2 SOLUTIONS: 13. Foam Produced in a Fire Extinguisher Reaction

Given Reaction (unbalanced):
NaHCO₃(s) + Al₂(SO₄)₃(s) → Al(OH)₃(s) + CO₂(g) + Na₂SO₄(s)

Step 1: Balance the Equation

• Count atoms on each side (starting with a trial balance):Try the following coefficients: 6 NaHCO₃ + Al₂(SO₄)₃ → 2 Al(OH)₃ + 6 CO₂ + 3 Na₂SO₄Check:
  • Na: 6 on left, 3*2=6 on right
  • Al: 2 on left, 2 on right
  • C: 6 (from 6 NaHCO₃) on left, 6 (from 6 CO₂) on right
  • H: 6 (from NaHCO₃) on left, 6 (from 2 Al(OH)₃, each with 3 H) on right
  • S: 3 (from Al₂(SO₄)₃) on left, 3 (from 3 Na₂SO₄) on right
  • O: 6*(3)=18 from NaHCO₃ plus 12 from Al₂(SO₄)₃ for a total of 30 O on left. On right:
    2 Al(OH)₃ = 2*(3 O)=6 O
    6 CO₂ = 6*(2 O)=12 O
    3 Na₂SO₄ = 3*(4 O)=12 O
    Total O right = 6+12+12=30 O. Perfect.

Balanced Equation:
6 NaHCO₃ + Al₂(SO₄)₃ → 2 Al(OH)₃ + 6 CO₂ + 3 Na₂SO₄

Step 2: Find Moles of NaHCO₃

• Given mass of NaHCO₃ = 1.000 kg = 1000 g
• Molar mass NaHCO₃: Na=23, H=1, C=12, O=3×16=48
  Total = 23+1+12+48 = 84 g/mol
• Moles NaHCO₃ = 1000 g / 84 g/mol ≈ 11.90 mol

Step 3: Use Stoichiometric Ratios

From the balanced equation:
6 moles NaHCO₃ produce 2 moles Al(OH)₃

Ratio: NaHCO₃ : Al(OH)₃ = 6 : 2 = 3 : 1

For 11.90 mol NaHCO₃, moles of Al(OH)₃ = 11.90 × (2/6) = 3.97 mol

Step 4: Convert Moles of Al(OH)₃ to Mass

Molar mass Al(OH)₃: Al=26.98, O=16×3=48, H=3×1=3
Total = 26.98 + 48 + 3 = 77.98 g/mol

Mass Al(OH)₃ = 3.97 mol × 77.98 g/mol ≈ 309.6 g

Answer #13: About 310 g of Al(OH)₃ foam is produced.

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14. Mass of Carbohydrate Consumed During Respiration

Reaction (respiration):
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

For every 1 mole C₆H₁₂O₆, 6 moles O₂ are required.

Step 1: Moles of O₂

Given 3.00 g O₂, M(O₂)=32 g/mol

Moles O₂ = 3.00 g / 32 g/mol = 0.09375 mol O₂

Step 2: Find Moles of Glucose (C₆H₁₂O₆)

Ratio: C₆H₁₂O₆ : O₂ = 1 : 6
Moles C₆H₁₂O₆ = 0.09375 / 6 = 0.015625 mol

Step 3: Mass of C₆H₁₂O₆

M(C₆H₁₂O₆) = 180 g/mol (C=12×6=72, H=12, O=16×6=96; total 72+12+96=180)

Mass = 0.015625 mol × 180 g/mol = 2.8125 g

Answer #14: About 2.81 g of carbohydrate are consumed per 3.00 g of O₂.

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15. Theoretical Yield of Copper from Zinc and Copper(II) Sulfate

Reaction:
Zn + CuSO₄ → Cu + ZnSO₄

Mole ratio: 1 Zn → 1 Cu

The problem states the theoretical yield of copper is 15.42 g and provides a final mass of zinc =14.15 g. The hint suggests determining how much Zn actually reacted (the initial mass of Zn minus final mass gives reacted Zn), then calculating Cu produced. However, the problem as given already states the theoretical yield is 15.42 g.

Answer #15: 15.42 g of copper (theoretical yield).

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16. Mass of Coal (C) Required to Produce Gasoline (C₈H₁₈)

Reaction (formation from elements):
8C (from coal) + 9H₂ → C₈H₁₈

Step 1: Moles of C₈H₁₈

Given 1.00 kg = 1000 g C₈H₁₈
M(C₈H₁₈): C=12×8=96, H=18, total=96+18=114 g/mol

Moles C₈H₁₈ = 1000 g / 114 g/mol ≈ 8.77 mol

Step 2: Moles of C Required

Ratio: C₈H₁₈ requires 8 moles C per 1 mole of C₈H₁₈

Moles C = 8.77 × 8 = 70.16 mol C

Step 3: Mass of C

M(C)=12 g/mol
Mass C = 70.16 mol × 12 g/mol ≈ 842 g

Answer #16: About 842 g of coal is needed.

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17. Moles of Cl₂ for Producing Bleaching Powder

Bleaching powder can be approximated by the reaction:
2 Ca(OH)₂ + 2 Cl₂ → Ca(OCl)₂ + CaCl₂ + 2 H₂O

From this reaction, for every 2 moles of Cl₂, we produce 1 mole Ca(OCl)₂ and 1 mole CaCl₂ together.

Molar Masses:

• Ca(OCl)₂: Ca=40.08, O=16, Cl=35.45
  OCl unit = O(16)+Cl(35.45)=51.45
  Ca(OCl)₂ = 40.08 + 2×51.45 = 40.08 +102.9 =142.98 g/mol
• CaCl₂: Ca=40.08 + 2×35.45=40.08+70.9=110.98 g/mol

Total mass of product pair from 2 moles Cl₂:
142.98 +110.98 =253.96 g per 2 moles Cl₂

Given: 1.000 kg (1000 g) product (the mixture Ca(OCl)₂ and CaCl₂)

Number of “sets” = 1000 g / 253.96 g ≈ 3.937 sets

Each set requires 2 moles Cl₂, so total Cl₂ = 3.937 × 2 = 7.874 moles Cl₂

Answer #17: About 7.87 moles of Cl₂ are required.

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18. Mass of NaOH Produced from Ca(OH)₂ and Na₂CO₃

Reaction:
Ca(OH)₂ + Na₂CO₃ → 2 NaOH + CaCO₃

Step 1: Moles of Ca(OH)₂

Given: 148 g Ca(OH)₂
M(Ca(OH)₂): Ca=40.08, O=16×2=32, H=2, total=40.08+32+2=74.08 g/mol

Moles Ca(OH)₂ =148 g /74.08 g/mol =2.0 mol

Step 2: Moles NaOH Produced

Ratio: 1 Ca(OH)₂ → 2 NaOH
From 2.0 mol Ca(OH)₂, we get 4.0 mol NaOH

Step 3: Mass of NaOH

M(NaOH)=Na(22.99)+O(16)+H(1)=39.99 g/mol ≈40.0 g/mol

Mass NaOH =4.0 mol ×40.0 g/mol=160 g

Answer #18: 160 g of NaOH can be produced.

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19. Moles of NO₂ from Reaction of Cu with HNO₃

Reaction:
Cu +4HNO₃ → Cu(NO₃)₂ +2NO₂ +2H₂O

Ratio: 1 Cu → 2 NO₂

Step 1: Moles of Cu

Given: 1.27 g Cu
M(Cu)=63.55 g/mol approximately

Moles Cu =1.27 g /63.55 g/mol ≈0.02 mol

Step 2: Moles NO₂

1 Cu →2 NO₂
From 0.02 mol Cu →0.04 mol NO₂

Answer #19: 0.0400 moles of NO₂.

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20. Mass of Water from H₂ and O₂ in a Rocket

Reaction:
2H₂ + O₂ → 2H₂O

Given: 5000 kg H₂ and 5000 kg O₂
Check limiting reagent.

Step 1: Moles of Each

• M(H₂)=2 g/mol
  5000 kg =5,000,000 g H₂
  Moles H₂=5,000,000 g /2 g/mol=2,500,000 mol H₂
• M(O₂)=32 g/mol
  5000 kg=5,000,000 g O₂
  Moles O₂=5,000,000 g /32 g/mol=156,250 mol O₂

Step 2: Determine Limiting Reagent

To consume all O₂ (156,250 mol), need double H₂: 2×156,250=312,500 mol H₂ required. We have 2,500,000 mol H₂, which is more than enough. O₂ is limiting.

Step 3: Moles H₂O Produced

O₂ : H₂O = 1 : 2
From 156,250 mol O₂ → 312,500 mol H₂O

Step 4: Mass of H₂O

M(H₂O)=18 g/mol
Mass H₂O =312,500 mol ×18 g/mol=5,625,000 g=5625 kg

Answer #20: 5625 kg of H₂O produced.

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21. Mass of Gypsum (CaSO₄·2H₂O) from Plaster of Paris ((CaSO₄)₂·H₂O)

Reaction Given:
(CaSO₄)₂·H₂O +3H₂O → 2 CaSO₄·2H₂O

Check ratio: 1 mole plaster →2 moles gypsum

Step 1: Molar Mass of (CaSO₄)₂·H₂O (Plaster)

M(CaSO₄)=136.14 g/mol (approx. Ca=40.08, S=32.06, O₄=64)
(CaSO₄)₂ =2×136.14=272.28 g
+H₂O=18 g
Total=272.28+18=290.28 g/mol

Step 2: Moles of Plaster

Given 29.04 g plaster
Moles=29.04 g /290.28 g/mol=0.1 mol

Step 3: Moles of Gypsum Produced

1 plaster →2 gypsum
From 0.1 mol plaster →0.2 mol gypsum (CaSO₄·2H₂O)

Step 4: Molar Mass Gypsum (CaSO₄·2H₂O)

CaSO₄=136.14 g/mol
2H₂O=36 g
Total=172.14 g/mol

Step 5: Mass of Gypsum

Mass=0.2 mol ×172.14 g/mol=34.428 g

Answer #21: About 34.4 g of gypsum is produced.

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All Answers:

13. ~310 g Al(OH)₃
14. ~2.81 g C₆H₁₂O₆
15. 15.42 g Cu
16. ~842 g C (coal)
17. ~7.87 mol Cl₂
18. 160 g NaOH
19. 0.0400 mol NO₂
20. 5625 kg H₂O
21. ~34.4 g CaSO₄·2H₂O (gypsum)

Mixed Problems

10. Electrolytic Decomposition of Water

Reaction:
2 H₂O → 2 H₂ + O₂

Ratio: 2 H₂ :1 O₂

Given: 0.500 mol H₂ formed
From ratio: O₂ =0.500 mol H₂ ×(1 O₂/2 H₂)=0.250 mol O₂

Answer #10: 0.250 mol O₂

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11. Neutralizing Stomach Acid with Milk of Magnesia

Reaction:
Mg(OH)₂ + 2 HCl → MgCl₂ + 2 H₂O

1 Mg(OH)₂ neutralizes 2 HCl.

Given: 0.583 g Mg(OH)₂
M(Mg(OH)₂)=24.31(Mg)+2×(16+1)=24.31+2×17=24.31+34=58.31 g/molmoles Mg(OH)2=0.583 g58.31 g/mol=0.01 mol\text{moles Mg(OH)}_2 = \frac{0.583 \, g}{58.31 \, g/mol} =0.01 \, \text{mol}moles Mg(OH)2​=58.31g/mol0.583g​=0.01mol

Moles HCl neutralized =0.01 mol Mg(OH)₂ ×2=0.02 mol HCl

Answer #11: 0.02 mol HCl

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12. Cream of Tartar and Baking Soda Reaction

Reaction:
KHC₄H₅O₆ + NaHCO₃ → CO₂ + H₂O + KNaC₄H₄O₆

Ratio: 1:1 between cream of tartar (KHC₄H₅O₆) and baking soda (NaHCO₃).

Given: 4.20 g NaHCO₃
M(NaHCO₃)=84 g/molmoles NaHCO3=4.20 g84 g/mol=0.05 mol\text{moles NaHCO}_3 = \frac{4.20 \, g}{84 \, g/mol}=0.05 \, \text{mol}moles NaHCO3​=84g/mol4.20g​=0.05mol

Moles of cream of tartar needed =0.05 mol (1:1 ratio)

M(KHC₄H₅O₆)=K(39.1) +H₅(5) +C₄(48) +O₆(96)
Sum=39.1+5+48+96=188.1 g/mol approx.

Mass cream of tartar=0.05 mol ×188.1 g/mol=9.41 g

Answer #12: 9.41 g KHC₄H₅O₆

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All Answers Recap:

Moles to Moles:
1(a): 0.250 mol S₈
1(b): 0.800 mol BaS
2(a): 8.00 mol O₂
2(b): 0.0800 mol H₂O
3(a): 0.200 mol P
3(b): 0.0833 mol Na₃P

Moles to Mass:
4: 24.8 g Na
5: 96.3 g CaCO₃

Mass to Moles:
6: 4.99 mol O₂
7: 0.20 mol HF

Mass to Mass:
8: ~226 g C per 1 kg Fe₂O₃
9: ~120 g Fe₂O₃ per 40.5 g Al

Mixed Problems:
10: 0.250 mol O₂
11: 0.02 mol HCl
12: 9.41 g KHC₄H₅O₆