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How many grams of Na are needed to react with
H₂O to liberate 4.00 x 102 mL of H₂ gas at STP?

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Answer:

The balanced chemical equation for the reaction of Na with H₂O is:

2Na + 2H₂O → 2NaOH + H₂

According to the given data, 4.00 x 10^2 mL of H₂ gas is produced at STP. We can use the ideal gas law to determine the number of moles of H₂ gas produced: PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

At STP, P = 1 atm and T = 273 K, so:

V = nRT/P = (1 mol)(0.0821 L atm/mol K)(273 K)/(1 atm) = 22.4 L

Therefore, 4.00 x 10^2 mL of H₂ gas is equal to 0.4 L.

We can use the stoichiometry of the balanced chemical equation to relate the moles of H₂ gas produced to the moles of Na required:

2 mol Na : 1 mol H₂

x mol Na : 0.5 mol H₂

x = 0.25 mol Na

The molar mass of Na is 22.99 g/mol, so:

0.25 mol Na x 22.99 g/mol = 5.75 g Na

Therefore, 5.75 grams of Na are needed to liberate 4.00 x 10^2 mL of H₂ gas at STP

User Narayan
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