Stoichiometry Review WS _1 - Foothill High School - Free Printable

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

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Problem 1:

Balancing the equation:

_Na + _O₂ → _Na₂O

#### Step 1: Balance the chemical equation

We have:
- Sodium (Na)
- Oxygen gas (O₂)
- Sodium oxide (Na₂O)

Unbalanced:
$$ \text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O} $$

Balance atoms:
- On the right, Na₂O has 2 Na and 1 O
- On the left, O₂ has 2 O, so we need 2 Na₂O to balance oxygen
- That gives us 4 Na on the right → need 4 Na on the left

So:
$$ 4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} $$

✔ Balanced equation:
$$ \boxed{4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}} $$

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#### a) How many moles of sodium (Na) would be needed to react with 3.82 moles of oxygen (O₂)?

From the balanced equation:
- 1 mole O₂ reacts with 4 moles Na

So:
$$
\text{Moles of Na} = 3.82 \, \text{mol O}_2 \times \frac{4 \, \text{mol Na}}{1 \, \text{mol O}_2} = 15.28 \, \text{mol Na}
$$

✔ Answer: 15.28 moles of Na

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#### b) How many moles of Na₂O can be produced from 13.5 moles of Na?

From the balanced equation:
- 4 moles Na → 2 moles Na₂O

So:
$$
\text{Moles of Na}_2\text{O} = 13.5 \, \text{mol Na} \times \frac{2 \, \text{mol Na}_2\text{O}}{4 \, \text{mol Na}} = 6.75 \, \text{mol Na}_2\text{O}
$$

✔ Answer: 6.75 moles of Na₂O

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#### c) How many molecules of O₂ are needed to produce 34.7 g of Na₂O?

Step 1: Molar mass of Na₂O
- Na: 23.0 g/mol × 2 = 46.0 g/mol
- O: 16.0 g/mol
- Total = 46.0 + 16.0 = 62.0 g/mol

Step 2: Moles of Na₂O
$$
\text{Moles} = \frac{34.7 \, \text{g}}{62.0 \, \text{g/mol}} = 0.5597 \, \text{mol} \approx 0.560 \, \text{mol}
$$

Step 3: Use stoichiometry
From balanced equation:
- 2 moles Na₂O ← 1 mole O₂
So:
$$
\text{Moles of O}_2 = 0.560 \, \text{mol Na}_2\text{O} \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol Na}_2\text{O}} = 0.280 \, \text{mol O}_2
$$

Step 4: Convert moles to molecules
Use Avogadro’s number: $ 6.022 \times 10^{23} $ molecules/mol

$$
\text{Molecules of O}_2 = 0.280 \times 6.022 \times 10^{23} = 1.686 \times 10^{23} \, \text{molecules}
$$

✔ Answer: $ 1.69 \times 10^{23} $ molecules of O₂ (rounded to 3 sig figs)

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Problem 2:

Balancing the equation:

_C₃H₈ + _O₂ → _CO₂ + _H₂O

Wait — the formula is written as C₃H₄, but that’s unusual. Let’s double-check.

Looking at the reaction: C₃H₄ is propyne or allene, but more likely this might be a typo. However, based on the structure and typical combustion problems, it’s possible it should be C₃H₈ (propane), but let's assume the problem says C₃H₄.

But let’s check: C₃H₄ is not common in standard combustion problems. But since it's given, we’ll proceed with C₃H₄.

So:
$$
\text{C}_3\text{H}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
$$

#### Step 1: Balance the equation

Unbalanced:
$$
\text{C}_3\text{H}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
$$

Balance C: 3 on left → 3 CO₂ on right
Balance H: 4 on left → 2 H₂O on right (since each has 2 H)

Now:
$$
\text{C}_3\text{H}_4 + \text{O}_2 \rightarrow 3\text{CO}_2 + 2\text{H}_2\text{O}
$$

Now count O:
- Right: 3×2 = 6 from CO₂, 2×1 = 2 from H₂O → total 8 O atoms
- So need 4 O₂ molecules (since each has 2 O)

Final balanced equation:
$$
\boxed{\text{C}_3\text{H}_4 + 4\text{O}_2 \rightarrow 3\text{CO}_2 + 2\text{H}_2\text{O}}
$$

✔ Balanced!

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#### a) When 0.624 moles of O₂ are reacted, how many moles of carbon dioxide are produced?

From balanced equation:
- 4 mol O₂ → 3 mol CO₂

So:
$$
\text{Moles of CO}_2 = 0.624 \, \text{mol O}_2 \times \frac{3 \, \text{mol CO}_2}{4 \, \text{mol O}_2} = 0.468 \, \text{mol CO}_2
$$

✔ Answer: 0.468 moles of CO₂

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#### b) How many grams of C₃H₄ are needed to produce 3.7 moles of water?

From balanced equation:
- 2 mol H₂O ← 1 mol C₃H₄

So:
$$
\text{Moles of C}_3\text{H}_4 = 3.7 \, \text{mol H}_2\text{O} \times \frac{1 \, \text{mol C}_3\text{H}_4}{2 \, \text{mol H}_2\text{O}} = 1.85 \, \text{mol C}_3\text{H}_4
$$

Now convert to grams:
Molar mass of C₃H₄:
- C: 12.0 × 3 = 36.0
- H: 1.0 × 4 = 4.0
- Total = 40.0 g/mol

$$
\text{Mass} = 1.85 \, \text{mol} \times 40.0 \, \text{g/mol} = 74.0 \, \text{g}
$$

✔ Answer: 74.0 grams of C₃H₄

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#### c) How many grams of O₂ are needed to react with 2.56 g of C₃H₄?

Step 1: Moles of C₃H₄
$$
\text{Moles} = \frac{2.56 \, \text{g}}{40.0 \, \text{g/mol}} = 0.0640 \, \text{mol}
$$

Step 2: Use stoichiometry
From balanced equation:
- 1 mol C₃H₄ → 4 mol O₂

So:
$$
\text{Moles of O}_2 = 0.0640 \times 4 = 0.256 \, \text{mol O}_2
$$

Step 3: Convert to grams
Molar mass O₂ = 32.0 g/mol

$$
\text{Mass} = 0.256 \, \text{mol} \times 32.0 \, \text{g/mol} = 8.192 \, \text{g} \approx 8.19 \, \text{g}
$$

✔ Answer: 8.19 grams of O₂

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✔ Final Answers Summary:

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#### Problem 1:
Balanced equation:
$$ 4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} $$

a) 15.28 moles of Na
b) 6.75 moles of Na₂O
c) $ 1.69 \times 10^{23} $ molecules of O₂

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#### Problem 2:
Balanced equation:
$$ \text{C}_3\text{H}_4 + 4\text{O}_2 \rightarrow 3\text{CO}_2 + 2\text{H}_2\text{O} $$

a) 0.468 moles of CO₂
b) 74.0 grams of C₃H₄
c) 8.19 grams of O₂

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