Homework WS4 Limiting Reactant Answerkey - Name: - Studocu - Free Printable

Let’s solve each problem step by step. We’ll use dimensional analysis (also called the factor-label method) to convert grams → moles → moles of product → grams of product. The key idea is: the reactant that produces the least amount of product is the limiting reactant, and that’s how much product we can actually make.

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Problem 1:

Given:
50.0 g KCl + 50.0 g O₂ → ? g KClO₃
Balanced equation:
2 KCl + 3 O₂ → 2 KClO₃

We need to find out how much KClO₃ each reactant can produce, then pick the smaller amount — that’s our answer.

#### Step 1: From KCl
Start with 50.0 g KCl.

→ Convert to moles of KCl:
Molar mass of KCl = 39.10 (K) + 35.45 (Cl) = 74.55 g/mol

So:
50.0 g KCl × (1 mol KCl / 74.55 g KCl) = 0.6707 mol KCl

→ Use mole ratio from balanced equation:
2 mol KCl → 2 mol KClO₃ → so ratio is 1:1

So: 0.6707 mol KCl → 0.6707 mol KClO₃

→ Convert to grams of KClO₃:
Molar mass of KClO₃ = 39.10 + 35.45 + 3×16.00 = 122.55 g/mol

0.6707 mol × 122.55 g/mol = 82.2 g KClO₃

✔ So from KCl, we get 82.2 g KClO₃

#### Step 2: From O₂
Start with 50.0 g O₂.

→ Molar mass of O₂ = 2×16.00 = 32.00 g/mol

50.0 g O₂ × (1 mol O₂ / 32.00 g O₂) = 1.5625 mol O₂

→ Mole ratio: 3 mol O₂ → 2 mol KClO₃

So: 1.5625 mol O₂ × (2 mol KClO₃ / 3 mol O₂) = 1.0417 mol KClO₃

→ Convert to grams:
1.0417 mol × 122.55 g/mol = 127.7 g ≈ 128 g KClO₃

✔ So from O₂, we get 128 g KClO₃

#### Step 3: Compare
KCl gives us 82.2 g, O₂ gives us 128 g.
The smaller amount wins → KCl is limiting.

✔ Final Answer for #1: 82.2 g KClO₃

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Problem 2:

Given:
34.5 g Cu + 70.2 g Ag(NO₃) → ? g Ag
Balanced equation:
Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2 Ag

Again, calculate how much Ag each reactant can make.

#### Step 1: From Cu
34.5 g Cu → moles?

Molar mass Cu = 63.55 g/mol (we’ll use 63.5 as in worksheet)

34.5 g × (1 mol Cu / 63.5 g) = 0.5433 mol Cu

→ Mole ratio: 1 mol Cu → 2 mol Ag

So: 0.5433 mol Cu × 2 = 1.0866 mol Ag

→ Grams of Ag: molar mass Ag = 107.87 g/mol

1.0866 mol × 107.87 g/mol = 117.2 g ≈ 117 g Ag

✔ From Cu: 117 g Ag

#### Step 2: From AgNO₃
70.2 g AgNO₃ → moles?

Molar mass AgNO₃ = 107.87 + 14.01 + 3×16.00 = 169.88 g/mol

70.2 g × (1 mol / 169.88 g) = 0.4132 mol AgNO₃

→ Mole ratio: 2 mol AgNO₃ → 2 mol Ag → so 1:1

So: 0.4132 mol AgNO₃ → 0.4132 mol Ag

→ Grams: 0.4132 mol × 107.87 g/mol = 44.57 g ≈ 44.6 g Ag

✔ From AgNO₃: 44.6 g Ag

#### Step 3: Compare
Cu makes 117 g, AgNO₃ makes 44.6 g → AgNO₃ is limiting

✔ Final Answer for #2: 44.6 g Ag

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Problem 3:

Given:
85 g NH₄Cl + 130 g Ca(OH)₂ → ? g NH₃
Balanced equation:
2 NH₄Cl + Ca(OH)₂ → CaCl₂ + 2 H₂O + 2 NH₃

Calculate NH₃ from each reactant.

#### Step 1: From NH₄Cl
85 g NH₄Cl → moles?

Molar mass NH₄Cl = 14.01 + 4×1.01 + 35.45 = 53.50 g/mol

85 g × (1 mol / 53.50 g) = 1.5888 mol NH₄Cl

→ Mole ratio: 2 mol NH₄Cl → 2 mol NH₃ → 1:1

So: 1.5888 mol NH₄Cl → 1.5888 mol NH₃

→ Grams NH₃: molar mass = 14.01 + 3×1.01 = 17.04 g/mol

1.5888 mol × 17.04 g/mol = 27.07 g ≈ 27 g NH₃

✔ From NH₄Cl: 27 g NH₃

#### Step 2: From Ca(OH)₂
130 g Ca(OH)₂ → moles?

Molar mass = 40.08 + 2×(16.00 + 1.01) = 40.08 + 34.02 = 74.10 g/mol

130 g × (1 mol / 74.10 g) = 1.7544 mol Ca(OH)₂

→ Mole ratio: 1 mol Ca(OH)₂ → 2 mol NH₃

So: 1.7544 mol × 2 = 3.5088 mol NH₃

→ Grams: 3.5088 mol × 17.04 g/mol = 59.79 g ≈ 60 g NH₃

Wait — but the worksheet says 6.0 × 10¹ g = 60 g, which matches.

But hold on — let’s check significant figures.

Original masses: 85 g (2 sig figs), 130 g (2 or 3? ambiguous, but likely 2).

In calculation from NH₄Cl: 85 g → 2 sig figs → answer should be 27 g (which has 2 sig figs).

From Ca(OH)₂: 130 g → if 2 sig figs, then 60 g (also 2 sig figs).

But since NH₄Cl gives only 27 g, and Ca(OH)₂ gives 60 g, the limiting reactant is NH₄Cl.

✔ Final Answer for #3: 27 g NH₃

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Final Answer:
1. 82.2 g KClO₃
2. 44.6 g Ag
3. 27 g NH₃