Stoichiometry practice problems worksheet focusing on converting between moles and grams using chemical equations.

Let’s solve each problem step by step. We’ll start with Problem 1, then move to Problem 2.

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Problem 1: Moles to Moles

We are given the unbalanced equation:

___C₂H₆ + ___O₂ → ___CO₂ + ___H₂O

First, we must balance it.

Step 1: Count atoms on each side.

Left: C=2, H=6, O=?
Right: C=1, H=2, O=3 (from CO₂ and H₂O)

We need to balance carbon first. Put a 2 in front of CO₂:

C₂H₆ + O₂ → 2CO₂ + H₂O

Now carbon is balanced (2 on each side).

Hydrogen: Left has 6, right has 2 → put a 3 in front of H₂O:

C₂H₆ + O₂ → 2CO₂ + 3H₂O

Now hydrogen is balanced (6 on each side).

Oxygen: Right side = 2×2 (from CO₂) + 3×1 (from H₂O) = 4 + 3 = 7 oxygen atoms.

Left side: O₂ → so we need 7/2 = 3.5 molecules of O₂.

But we can’t have fractions in chemical equations — multiply entire equation by 2:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

✔ Balanced equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Now use mole ratios from this balanced equation.

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Part a) If 4.50 moles of C₂H₆ are used, how many moles of oxygen are required?

From balanced equation:
2 moles C₂H₆ require 7 moles O₂

So set up ratio:

(7 moles O₂ / 2 moles C₂H₆) × 4.50 moles C₂H₆

= (7/2) × 4.50
= 3.5 × 4.50
= 15.75 moles O₂

✔ Answer for part a): 15.75 moles of oxygen

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Part b) If 0.3 moles of H₂O were produced, how many moles of carbon dioxide would also have been produced?

From balanced equation:
6 moles H₂O are produced along with 4 moles CO₂

So ratio: 4 moles CO₂ : 6 moles H₂O → simplify to 2:3

So for every 3 moles H₂O, you get 2 moles CO₂.

Set up:

(4 moles CO₂ / 6 moles H₂O) × 0.3 moles H₂O
= (2/3) × 0.3
= 0.2 moles CO₂

✔ Answer for part b): 0.2 moles of carbon dioxide

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Part c) If you wanted to produce 5 moles of carbon dioxide, how many moles of oxygen would you needed to have started with?

From balanced equation:
4 moles CO₂ come from 7 moles O₂

So ratio: 7 moles O₂ / 4 moles CO₂

Multiply by 5 moles CO₂:

(7/4) × 5 = 35/4 = 8.75 moles O₂

✔ Answer for part c): 8.75 moles of oxygen

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Problem 2: Grams to Moles

Unbalanced equation:

___H₂SO₄ + ___NaOH → ___Na₂SO₄ + ___H₂O

Balance it.

Left: H₂SO₄ has 2H, 1S, 4O; NaOH has 1Na, 1O, 1H
Right: Na₂SO₄ has 2Na, 1S, 4O; H₂O has 2H, 1O

Start with sodium: Right has 2 Na in Na₂SO₄ → need 2 NaOH on left.

Try:

H₂SO₄ + 2NaOH → Na₂SO₄ + H₂O

Check atoms:

Left: H = 2 (from H₂SO₄) + 2 (from 2NaOH) = 4H
S = 1, O = 4 + 2 = 6, Na = 2

Right: Na₂SO₄ → 2Na, 1S, 4O; H₂O → 2H, 1O → total H=2, O=5 → not balanced!

Wait — water should be 2H₂O to match hydrogens.

Try:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Now check:

Left: H = 2 + 2×1 = 4; S=1; O=4+2=6; Na=2
Right: Na₂SO₄ → 2Na, 1S, 4O; 2H₂O → 4H, 2O → total H=4, O=6, Na=2, S=1 ✔

Balanced equation:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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Part a) How many grams of H₂SO₄ are needed to react with 0.75 moles of NaOH?

From balanced equation:
2 moles NaOH react with 1 mole H₂SO₄

So moles of H₂SO₄ needed = (1/2) × 0.75 = 0.375 moles

Now convert moles to grams.

Molar mass of H₂SO₄:
H = 1 × 2 = 2
S = 32
O = 16 × 4 = 64
Total = 2 + 32 + 64 = 98 g/mol

Grams = moles × molar mass = 0.375 × 98

Calculate:
0.375 × 100 = 37.5 → but 98 is 2 less than 100 → subtract 0.375×2 = 0.75
→ 37.5 - 0.75 = 36.75

Or directly:
0.375 × 98 = (3/8) × 98 = (3 × 98)/8 = 294 / 8 = 36.75

✔ Answer for part a): 36.75 grams of H₂SO₄

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Part b) How many grams of Na₂SO₄ would be formed if 0.75 moles of NaOH reacted?

From balanced equation:
2 moles NaOH produce 1 mole Na₂SO₄

So moles of Na₂SO₄ = (1/2) × 0.75 = 0.375 moles

Molar mass of Na₂SO₄:
Na = 23 × 2 = 46
S = 32
O = 16 × 4 = 64
Total = 46 + 32 + 64 = 142 g/mol

Grams = 0.375 × 142

Calculate:
0.375 × 140 = 52.5
0.375 × 2 = 0.75
Total = 52.5 + 0.75 = 53.25

Or: (3/8) × 142 = (3 × 142)/8 = 426 / 8 = 53.25

✔ Answer for part b): 53.25 grams of Na₂SO₄

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Final Answer:
Problem 1a: 15.75 moles of oxygen
Problem 1b: 0.2 moles of carbon dioxide
Problem 1c: 8.75 moles of oxygen
Problem 2a: 36.75 grams of H₂SO₄
Problem 2b: 53.25 grams of Na₂SO₄