Question #764c2

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Answer 1 · 2015-04-19T16:35:29

The volume of hydrogen is 57.4 L.

We start with the balanced equation.

$2Al + {6HNO}_{3} \to {2Al(NO}_{3})_{3} + {3H}_{2}$ $"2Al" + "6HNO"_3 → "2Al(NO"_3")"_3 + "3H"_2$

Step 1. Calculate the moles of $Al$ $"Al"$ .

$45.3 cancel("g Al") × "1 mol Al"/(26.98 cancel("g Al")) = "1.679 mol Al"$

Step 2. Calculate the moles of $"H"_2$ .

The balanced equation tells us that 2 mol of $"Al"$ give 3 mol of $"H"_2$ . So,

$1.679 cancel("mol Al") × ("3 mol H"_2)/(2 cancel("mol Al")) = "2.519 mol H"_2$

Step3. To calculate the volume, we use the Ideal Gas Law:

$PV = nRT$

$V = (nRT)/P = (2.519 cancel("mol") × "0.082 06 L"·cancel("atm·K⁻¹mol⁻¹") × 300.15 cancel("K"))/(1.08 cancel("atm")) = "57.4 L"$