Worksheet titled "Mass to Mass Stoichiometry Worksheet" featuring four problems that require balancing chemical equations and calculating masses of reactants and products.

Let's solve each problem step-by-step using mass-to-mass stoichiometry, which involves converting mass to moles, using mole ratios from balanced equations, and then converting back to mass.

---

Problem 1:

Balanced Reaction:
$$
\text{Li}_3\text{N}(s) + 3\text{H}_2\text{O}(l) \rightarrow \text{NH}_3(g) + 3\text{LiOH}(aq)
$$

#### a. What mass of lithium hydroxide is produced when 0.38 g of lithium nitride react?

Step 1: Molar masses
- $\text{Li}_3\text{N}$: $3(6.94) + 14.01 = 20.82 + 14.01 = 34.83\ \text{g/mol}$
- $\text{LiOH}$: $6.94 + 16.00 + 1.01 = 23.95\ \text{g/mol}$

Step 2: Moles of Li₃N
$$
\text{moles of Li}_3\text{N} = \frac{0.38\ \text{g}}{34.83\ \text{g/mol}} = 0.0110\ \text{mol}
$$

Step 3: Use mole ratio
From the balanced equation:
$$
1\ \text{mol Li}_3\text{N} \rightarrow 3\ \text{mol LiOH}
$$
So,
$$
\text{moles of LiOH} = 0.0110 \times 3 = 0.0330\ \text{mol}
$$

Step 4: Convert to mass
$$
\text{mass of LiOH} = 0.0330\ \text{mol} \times 23.95\ \text{g/mol} = 0.790\ \text{g}
$$

✔ Answer a: 0.790 g of LiOH

---

#### b. How many grams of lithium nitride would react with 4.05 g of H₂O?

Step 1: Molar mass of H₂O
- $2(1.01) + 16.00 = 18.02\ \text{g/mol}$

Step 2: Moles of H₂O
$$
\frac{4.05\ \text{g}}{18.02\ \text{g/mol}} = 0.2247\ \text{mol}
$$

Step 3: Mole ratio
From equation: $3\ \text{mol H}_2\text{O} \leftrightarrow 1\ \text{mol Li}_3\text{N}$

So,
$$
\text{moles of Li}_3\text{N} = \frac{0.2247}{3} = 0.0749\ \text{mol}
$$

Step 4: Mass of Li₃N
$$
0.0749\ \text{mol} \times 34.83\ \text{g/mol} = 2.57\ \text{g}
$$

✔ Answer b: 2.57 g of Li₃N

---

Problem 2:

Unbalanced reaction:
$$
2\text{NaI}(s) + \text{Cl}_2(g) \rightarrow 2\text{NaCl}(s) + \text{I}_2(g)
$$
This is already balanced.

#### a. What mass of sodium chloride is produced when 0.294 g of sodium iodide react?

Molar masses:
- NaI: $22.99 + 126.90 = 149.89\ \text{g/mol}$
- NaCl: $22.99 + 35.45 = 58.44\ \text{g/mol}$

Step 1: Moles of NaI
$$
\frac{0.294\ \text{g}}{149.89\ \text{g/mol}} = 0.001961\ \text{mol}
$$

Step 2: Mole ratio
$2\ \text{mol NaI} \rightarrow 2\ \text{mol NaCl}$ → 1:1 ratio

So, moles of NaCl = 0.001961 mol

Step 3: Mass of NaCl
$$
0.001961 \times 58.44 = 0.1145\ \text{g} \approx 0.115\ \text{g}
$$

✔ Answer a: 0.115 g of NaCl

---

#### b. If 5.80 g of iodine is formed, what is the mass of sodium iodide that reacted?

Molar mass I₂: $2(126.90) = 253.80\ \text{g/mol}$

Step 1: Moles of I₂
$$
\frac{5.80\ \text{g}}{253.80\ \text{g/mol}} = 0.02285\ \text{mol}
$$

Step 2: Mole ratio
From equation: $1\ \text{mol I}_2 \leftarrow 2\ \text{mol NaI}$

So,
$$
\text{moles of NaI} = 2 \times 0.02285 = 0.04570\ \text{mol}
$$

Step 3: Mass of NaI
$$
0.04570 \times 149.89 = 6.848\ \text{g} \approx 6.85\ \text{g}
$$

✔ Answer b: 6.85 g of NaI

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Problem 3: Combustion of butane (C₄H₁₀)

Step 1: Write and balance the combustion reaction

$$
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
$$

Balance:
- C: 4 → 4 CO₂
- H: 10 → 5 H₂O
- O: Right side: $4(2) + 5(1) = 8 + 5 = 13$ → So need $13/2$ O₂

Multiply entire equation by 2:
$$
2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}
$$

Now balanced.

Given: 54.5 g of butane (C₄H₁₀)

Molar masses:
- C₄H₁₀: $4(12.01) + 10(1.01) = 48.04 + 10.10 = 58.14\ \text{g/mol}$
- CO₂: $12.01 + 2(16.00) = 44.01\ \text{g/mol}$

Step 1: Moles of C₄H₁₀
$$
\frac{54.5\ \text{g}}{58.14\ \text{g/mol}} = 0.9375\ \text{mol}
$$

Step 2: Mole ratio
From equation: $2\ \text{mol C}_4\text{H}_{10} \rightarrow 8\ \text{mol CO}_2$

So,
$$
\text{moles of CO}_2 = 0.9375 \times \frac{8}{2} = 0.9375 \times 4 = 3.75\ \text{mol}
$$

Step 3: Mass of CO₂
$$
3.75\ \text{mol} \times 44.01\ \text{g/mol} = 165.0\ \text{g}
$$

✔ Answer: 165 g of CO₂ are produced

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Problem 4:

Unbalanced reaction:
$$
4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
$$
This is already balanced.

#### a. How many grams of iron(IV) sulfide are used when 9 g of O₂ react?

Wait — iron(IV) sulfide is FeS₂? Actually, FeS₂ is iron(II) disulfide, not iron(IV). But in this context, it’s commonly referred to as "pyrite" and the formula is correct.

But let’s proceed.

Molar masses:
- O₂: $32.00\ \text{g/mol}$
- FeS₂: $55.85 + 2(32.07) = 55.85 + 64.14 = 119.99\ \text{g/mol}$

Step 1: Moles of O₂
$$
\frac{9\ \text{g}}{32.00\ \text{g/mol}} = 0.28125\ \text{mol}
$$

Step 2: Mole ratio
From equation: $11\ \text{mol O}_2 \leftrightarrow 4\ \text{mol FeS}_2$

So,
$$
\text{moles of FeS}_2 = 0.28125 \times \frac{4}{11} = 0.1010\ \text{mol}
$$

Step 3: Mass of FeS₂
$$
0.1010 \times 119.99 = 12.12\ \text{g}
$$

✔ Answer a: 12.1 g of FeS₂

---

#### b. How much iron(III) oxide is produced when 25 g of iron(IV) sulfide are used?

Wait — iron(III) oxide is Fe₂O₃ (correct), and iron(IV) sulfide is likely a misnomer — FeS₂ is actually iron(II) sulfide. But we'll use the given formula.

We’re using FeS₂ (molar mass ≈ 119.99 g/mol)

Step 1: Moles of FeS₂
$$
\frac{25\ \text{g}}{119.99\ \text{g/mol}} = 0.2083\ \text{mol}
$$

Step 2: Mole ratio
From equation: $4\ \text{mol FeS}_2 \rightarrow 2\ \text{mol Fe}_2\text{O}_3$

So,
$$
\text{moles of Fe}_2\text{O}_3 = 0.2083 \times \frac{2}{4} = 0.10415\ \text{mol}
$$

Step 3: Molar mass of Fe₂O₃
- $2(55.85) + 3(16.00) = 111.7 + 48.00 = 159.7\ \text{g/mol}$

Mass of Fe₂O₃:
$$
0.10415 \times 159.7 = 16.63\ \text{g}
$$

✔ Answer b: 16.6 g of Fe₂O₃

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✔ Final Answers:

#### 1a. 0.790 g LiOH
#### 1b. 2.57 g Li₃N
#### 2a. 0.115 g NaCl
#### 2b. 6.85 g NaI
#### 3. 165 g CO₂
#### 4a. 12.1 g FeS₂
#### 4b. 16.6 g Fe₂O₃

Let me know if you'd like these rounded differently or want explanations in a different format!