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.

Let's break down the problem step by step. ### 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