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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?

User Ria
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1 Answer

4 votes

Answer:

6.57 L

Step-by-step 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} * \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} * \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 = (nRT )/( p)


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

User Veilig
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