Questions: How many grams of Na solid are required to completely react with 19.2 L of Cl2 gas at STP according to the following chemical reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP. 2 Na(s) + Cl2(g) -> 2 NaCl(s)

Solution Steps

At standard temperature and pressure (STP), 1 mole of an ideal gas occupies 22.4 liters. Given that we have 19.2 liters of \(\mathrm{Cl}_2\) gas, we can calculate the number of moles of \(\mathrm{Cl}_2\) using the formula:

\[ \text{Moles of } \mathrm{Cl}_2 = \frac{\text{Volume of } \mathrm{Cl}_2}{\text{Volume of 1 mole at STP}} = \frac{19.2 \, \text{L}}{22.4 \, \text{L/mol}} \]

\[ \text{Moles of } \mathrm{Cl}_2 = 0.8571 \, \text{mol} \]

The balanced chemical equation is:

\[ 2 \mathrm{Na}(\mathrm{s}) + \mathrm{Cl}_2(\mathrm{g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s}) \]

From the equation, 1 mole of \(\mathrm{Cl}_2\) reacts with 2 moles of \(\mathrm{Na}\). Therefore, the moles of \(\mathrm{Na}\) required are:

\[ \text{Moles of } \mathrm{Na} = 2 \times \text{Moles of } \mathrm{Cl}_2 = 2 \times 0.8571 \, \text{mol} = 1.7142 \, \text{mol} \]

The molar mass of sodium (\(\mathrm{Na}\)) is approximately 22.99 g/mol. Therefore, the mass of \(\mathrm{Na}\) required is:

\[ \text{Mass of } \mathrm{Na} = \text{Moles of } \mathrm{Na} \times \text{Molar mass of } \mathrm{Na} = 1.7142 \, \text{mol} \times 22.99 \, \text{g/mol} \]

\[ \text{Mass of } \mathrm{Na} = 39.42 \, \text{g} \]

Final Answer