Question

if 15.8 grams of sodium react with excess water, how many liters of hydrogen gas can be produced at 303 Kelvin at 1.30 atmospheres?

Answer 1

Answer:

6.57 L

Explanation:

First, calculate the moles of hydrogen produced, then use the Ideal Gas Law to calculate the volume of hydrogen.

Step 1. Write the chemical equation.

$M_{r}$: 22.99

           2Na + H₂O ⟶ 2NaOH + H₂

Step 1. Convert grams of Na to moles of Na

$\text{Moles of Na} = \text{15.8 g Na} \times \frac{\text{1mol Na} }{\text{22.99 g Na}}=\text{0.6873 mol Na}$

Step 2. Use the molar ratio of H₂:Na to convert moles of Na to moles of H₂.

$\text{Moles of H}_{2} = \text{0.6873 mol Na} \times \frac{\text{1 mol H}_{2}}{\text{2 mol Na} } = \text{0.3436 mol H}_{2}$

Step 3. Use the Ideal Gas Law to calculate the volume of hydrogen.

pV = nRT

$V = \frac{nRT }{ p}$

$V = \frac{\text{0.3436 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1}\times \text{303 K}}{\text{1.30 atm}} = \textbf{6.57 L}$