Molar mass of lithium nitride would react with 4.05 g of $$ \mathrm{H}_{2} \mathrm{O} $$ ? 2. Balance the following reaction and answer the given questions. $$ 2 \mathrm{Nal}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})+\mathrm{I}_{2}(\mathrm{~g}) $$ a. What mass of sodium chloride is produced when 0.294 g of sodium iodide react? b. If 5.80 g of iodine is formed, what is the mass of sodium iodide that reacted? 3. In the combustion of 54.5 g of butane $$ \left(\mathrm{C}_{4} \mathrm{H}_{6}\right) $$, how many grams of $$ \mathrm{CO}_{2} $$ are produced? Write and balance the equation before solving.

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### 1. Molar Mass of Lithium Nitride and Reaction with Water

The molar mass of lithium nitride ($$ \mathrm{Li}_{3}\mathrm{N} $$) can be calculated as follows:
- Lithium (Li) has a molar mass of approximately 6.94 g/mol.
- Nitrogen (N) has a molar mass of approximately 14.01 g/mol.

The molar mass of lithium nitride is:
$$
\text{Molar mass of } \mathrm{Li}_{3}\mathrm{N} = 3 \times 6.94 + 14.01 = 20.82 + 14.01 = 34.83 \text{ g/mol}
$$

### 2. Balancing the Reaction

The given reaction is:
$$
2 \mathrm{NaI}(\mathrm{s}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s}) + \mathrm{I}_{2}(\mathrm{g})
$$
This reaction is already balanced.

#### a. Mass of Sodium Chloride Produced from Sodium Iodide

From the balanced equation, we see that:
- 2 moles of $$ \mathrm{NaI} $$ produce 2 moles of $$ \mathrm{NaCl} $$.

This means the molar ratio of $$ \mathrm{NaI} $$ to $$ \mathrm{NaCl} $$ is 1:1.

First, we need to calculate the number of moles of $$ \mathrm{NaI} $$ in 0.294 g:
- Molar mass of $$ \mathrm{NaI} $$ (Sodium Iodide):
  - Sodium (Na) = 22.99 g/mol
  - Iodine (I) = 126.90 g/mol
$$
\text{Molar mass of } \mathrm{NaI} = 22.99 + 126.90 = 149.89 \text{ g/mol}
$$

Now, calculate the moles of $$ \mathrm{NaI} $$:
$$
\text{Moles of } \mathrm{NaI} = \frac{0.294 \text{ g}}{149.89 \text{ g/mol}} \approx 0.00196 \text{ mol}
$$

Since the molar ratio of $$ \mathrm{NaI} $$ to $$ \mathrm{NaCl} $$ is 1:1, the moles of $$ \mathrm{NaCl} $$ produced will also be 0.00196 mol.

Now, calculate the mass of $$ \mathrm{NaCl} $$:
- Molar mass of $$ \mathrm{NaCl} $$:
$$
\text{Molar mass of } \mathrm{NaCl} = 22.99 + 35.45 = 58.44 \text{ g/mol}
$$
$$
\text{Mass of } \mathrm{NaCl} = 0.00196 \text{ mol} \times 58.44 \text{ g/mol} \approx 0.114 \text{ g}
$$

#### b. Mass of Sodium Iodide that Reacted with Iodine

From the balanced equation, we see that:
- 2 moles of $$ \mathrm{NaI} $$ produce 1 mole of $$ \mathrm{I}_{2} $$.

Given that 5.80 g of $$ \mathrm{I}_{2} $$ is formed, we first calculate the moles of $$ \mathrm{I}_{2} $$:
- Molar mass of $$ \mathrm{I}_{2} $$:
$$
\text{Molar mass of } \mathrm{I}_{2} = 2 \times 126.90 = 253.80 \text{ g/mol}
$$
$$
\text{Moles of } \mathrm{I}_{2} = \frac{5.80 \text{ g}}{253.80 \text{ g/mol}} \approx 0.0228 \text{ mol}
$$

Since the molar ratio of $$ \mathrm{NaI} $$ to $$ \mathrm{I}_{2} $$ is 2:1, the moles of $$ \mathrm{NaI} $$ that reacted is:
$$
\text{Moles of } \mathrm{NaI} = 2 \times 0.0228 \text{ mol} = 0.0456 \text{ mol}
$$

Now, calculate the mass of $$ \mathrm{NaI} $$:
$$
\text{Mass of } \mathrm{NaI} = 0.0456 \text{ mol} \times 149.89 \text{ g/mol} \approx 6.83 \text{ g}
$$

### 3. Combustion of Butane

The combustion reaction for butane ($$ \mathrm{C}_{4}\mathrm{H}_{10} $$) is:
$$
\mathrm{C}_{4}\mathrm{H}_{10} + 6 \mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2} + 5 \mathrm{H}_{2}\mathrm{O}
$$

Now, we need to balance the equation:
- 1 mole of butane reacts with 6 moles of oxygen to produce 4 moles of carbon dioxide and 5 moles of water.

Next, we calculate the moles of butane in 54.5 g:
- Molar mass of $$ \mathrm{C}_{4}\mathrm{H}_{10} $$:
$$
\text{Molar mass of } \mathrm{C}_{4}\mathrm{H}_{10} = 4 \times 12.01 + 10 \times 1.008 = 48.04 + 10.08 = 58.12 \text{ g/mol}
$$
$$
\text{Moles of } \mathrm{C}_{4}\mathrm{H}_{10} = \frac{54.5 \text{ g}}{58.12 \text{ g/mol}} \approx 0.937 \text{ mol}
$$

From the balanced equation, 1 mole of butane produces 4 moles of $$ \mathrm{CO}_{2} $$:
$$
\text{Moles of } \mathrm{CO}_{2} = 0.937 \text{ mol} \times 4 = 3.748 \text{ mol}
$$

Now, calculate the mass of $$ \mathrm{CO}_{2} $$:
- Molar mass of $$ \mathrm{CO}_{2} $$:
$$
\text{Molar mass of } \mathrm{CO}_{2} = 12.01 + 2 \times 16.00 = 12.01 + 32.00 = 44.01 \text{ g/mol}
$$
\[
\text{Mass of } \mathrm{CO}_{2} = 3.748
**1. Molar Mass of Lithium Nitride:** - Lithium nitride ($$ \mathrm{Li}_{3}\mathrm{N} $$) has a molar mass of approximately 34.83 g/mol. **2. Reaction of Sodium Iodide with Chlorine Gas:** - **a. Mass of Sodium Chloride Produced:** - 0.294 g of $$ \mathrm{NaI} $$ produces approximately 0.114 g of $$ \mathrm{NaCl} $$. - **b. Mass of Sodium Iodide Reacted with Iodine:** - 5.80 g of $$ \mathrm{I}_{2} $$ requires approximately 6.83 g of $$ \mathrm{NaI} $$. **3. Combustion of Butane:** - 54.5 g of butane ($$ \mathrm{C}_{4}\mathrm{H}_{10} $$) produces approximately 3.748 g of $$ \mathrm{CO}_{2} $$.

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