Chemistry worksheet with stoichiometry practice problems.

Let’s solve each problem step by step. We’ll use balanced chemical equations and mole ratios to find the answers.

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Problem 1:
> If 20.0 g of magnesium react with excess hydrochloric acid, how many grams of magnesium chloride will be produced?
> Mg(s) + HCl(aq) → MgCl₂(aq) + H₂(g)

First, balance the equation:
Mg + 2HCl → MgCl₂ + H₂

Now, molar masses:
- Mg = 24.3 g/mol
- MgCl₂ = 24.3 + (2 × 35.5) = 95.3 g/mol

Step 1: Moles of Mg = mass / molar mass = 20.0 g / 24.3 g/mol ≈ 0.823 mol

Step 2: From the balanced equation, 1 mol Mg → 1 mol MgCl₂
So, moles of MgCl₂ = 0.823 mol

Step 3: Mass of MgCl₂ = moles × molar mass = 0.823 mol × 95.3 g/mol ≈ 78.4 g

✔ Check: 20.0 g Mg → ~78.4 g MgCl₂ — makes sense since MgCl₂ is heavier per mole.

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Problem 2:
> How many grams of aluminum are required to produce 550.0 g of Fe using the following reaction?
> Al + Fe₃O → Fe + Al₂O

Balance the equation first:

We need to balance atoms on both sides.

Try:
8Al + 3Fe₃O₄ → 9Fe + 4Al₂O₃

Check:
Left: Al=8, Fe=9, O=12
Right: Fe=9, Al=8, O=12 → Balanced!

Molar masses:
- Fe = 55.8 g/mol
- Al = 27.0 g/mol

Step 1: Moles of Fe desired = 550.0 g / 55.8 g/mol ≈ 9.8566 mol

Step 2: From equation, 9 mol Fe ← 8 mol Al
So, moles of Al needed = (8/9) × 9.8566 ≈ 8.7614 mol

Step 3: Mass of Al = 8.7614 mol × 27.0 g/mol ≈ 236.6 g

✔ Check: To make 550g Fe (~9.86 mol), we need about 8.76 mol Al → 236.6g — reasonable.

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Problem 3:
> If excess ammonium sulfate reacts with 20.0 g of calcium hydroxide, how many grams of ammonia gas are produced?
> (NH₄)₂SO₄(aq) + Ca(OH)₂(aq) → CaSO₄(s) + NH₃(g) + H₂O(l)

Balance the equation:

(NH₄)₂SO₄ + Ca(OH)₂ → CaSO₄ + 2NH₃ + 2H₂O

Check:
Left: N=2, H=8+2=10, S=1, O=4+2=6, Ca=1
Right: Ca=1, S=1, O=4+2=6, N=2, H=6+4=10 → Balanced!

Molar masses:
- Ca(OH)₂ = 40.1 + 2×(16.0 + 1.0) = 74.1 g/mol
- NH₃ = 14.0 + 3×1.0 = 17.0 g/mol

Step 1: Moles of Ca(OH)₂ = 20.0 g / 74.1 g/mol ≈ 0.2699 mol

Step 2: From equation, 1 mol Ca(OH)₂ → 2 mol NH₃
So, moles of NH₃ = 2 × 0.2699 ≈ 0.5398 mol

Step 3: Mass of NH₃ = 0.5398 mol × 17.0 g/mol ≈ 9.18 g

✔ Check: 20g Ca(OH)₂ gives about 9.18g NH₃ — seems right.

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Problem 4:
> How many grams of sodium sulfate will be formed if you start with 25.0 g of sulfuric acid in the following reaction?
> NaOH + H₂SO₄ → Na₂SO₄ + H₂O

Balance the equation:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Molar masses:
- H₂SO₄ = 2×1.0 + 32.1 + 4×16.0 = 98.1 g/mol
- Na₂SO₄ = 2×23.0 + 32.1 + 4×16.0 = 142.1 g/mol

Step 1: Moles of H₂SO₄ = 25.0 g / 98.1 g/mol ≈ 0.2548 mol

Step 2: From equation, 1 mol H₂SO₄ → 1 mol Na₂SO₄
So, moles of Na₂SO₄ = 0.2548 mol

Step 3: Mass of Na₂SO₄ = 0.2548 mol × 142.1 g/mol ≈ 36.2 g

✔ Check: 25g acid → ~36g salt — logical since Na₂SO₄ is heavier than H₂SO₄ per mole.

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Problem 5:
> What mass of oxygen is needed to completely react with 8.0 g of CO according to the following reaction?
> CO + O₂ → CO₂

Balance the equation:

2CO + O₂ → 2CO₂

Molar masses:
- CO = 12.0 + 16.0 = 28.0 g/mol
- O₂ = 32.0 g/mol

Step 1: Moles of CO = 8.0 g / 28.0 g/mol ≈ 0.2857 mol

Step 2: From equation, 2 mol CO require 1 mol O₂
So, moles of O₂ needed = 0.2857 / 2 = 0.14285 mol

Step 3: Mass of O₂ = 0.14285 mol × 32.0 g/mol ≈ 4.57 g

✔ Check: 8g CO needs half its moles in O₂ → 4.57g — correct.

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Problem 6:
> Sodium metal reacts with water to produce sodium hydroxide. What mass of sodium metal must be reacted to form 10.0 grams of sodium hydroxide?

Write and balance the equation:

2Na + 2H₂O → 2NaOH + H₂

Simplify: Na + H₂O → NaOH + ½H₂ — but better to keep whole numbers.

Use: 2Na + 2H₂O → 2NaOH + H₂

Molar masses:
- NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol
- Na = 23.0 g/mol

Step 1: Moles of NaOH desired = 10.0 g / 40.0 g/mol = 0.250 mol

Step 2: From equation, 2 mol NaOH ← 2 mol Na → so 1:1 ratio
Thus, moles of Na needed = 0.250 mol

Step 3: Mass of Na = 0.250 mol × 23.0 g/mol = 5.75 g

✔ Check: 10g NaOH requires 5.75g Na — makes sense because Na is lighter than NaOH.

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Problem 7:
> When hexane, C₆H₁₄, burns in oxygen, the products are carbon dioxide and water. What mass of carbon dioxide will be formed when 15.0 grams of hexane is burned in excess oxygen?

Write and balance the combustion reaction:

C₆H₁ + O₂ → CO₂ + H₂O

Balance:

C₆H₁₄ + 19/2 O₂ → 6CO₂ + 7H₂O
Or multiply by 2:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

But for calculation, we can use:
C₆H₁₄ + 9.5O₂ → 6CO₂ + 7H₂O

Molar masses:
- C₆H₁ = 6×12.0 + 14×1.0 = 86.0 g/mol
- CO₂ = 12.0 + 2×16.0 = 44.0 g/mol

Step 1: Moles of C₆H₁₄ = 15.0 g / 86.0 g/mol ≈ 0.1744 mol

Step 2: From equation, 1 mol C₆H₁₄ → 6 mol CO₂
So, moles of CO₂ = 6 × 0.1744 ≈ 1.0464 mol

Step 3: Mass of CO₂ = 1.0464 mol × 44.0 g/mol ≈ 46.0 g

✔ Check: 15g hexane → ~46g CO₂ — reasonable since CO₂ has more mass per carbon atom.

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Final Answer:

1. 78.4 g
2. 236.6 g
3. 9.18 g
4. 36.2 g
5. 4.57 g
6. 5.75 g
7. 46.0 g