Question #97b37

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Question #97b37

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Answer 1 · 2015-03-13T15:00:17

You can form 9.20 L of NH₃ gas at 93.0 °C and 37.3 kPa or 2.56 L at STP.

The balanced chemical equation is

N₂ + 3H₂ → 2NH₃

Since all substances are gases, we can use litres instead of moles in our calculations.

You don't specify the pressure and temperature of the NH₃, so I will calculate its volume in two ways.

At 93.0 °C and 37.3 kPa

${13.8 L H}_{2} \times \frac{{2 L NH}_{3}}{{3 L H}_{2}} = {9.20 L NH}_{3}$ $"13.8 L H"_2 × "2 L NH"_3/"3 L H"_2 = "9.20 L NH"_3$

At STP

STP is 100 kPa and 273.15 K

We can use the combined gas laws to calculate the volume at STP.

$P_{1} = 37.3 kPa; V_{1} = 9.20 L; T_{1} = 93.0 °C = 366.15 K$ $P_1 = "37.3 kPa"; V_1 = "9.20 L"; T_1 = "93.0 °C" = "366.15 K"$
$P_{2} = 100 kPa; V_{2} = ?; T_{2} = 273.15 K$ $P_2 = "100 kPa"; V_2 = "?"; T_2 = "273.15 K"$

$V_{2} = V_{1} \times Pressure factor \times Temperature factor$ $V_2 = V_1 × "Pressure factor" × "Temperature factor"$

$V_{2} = 9.20 L \times \frac{37.3 kPa}{100 kPa} \times \frac{273.15 K}{366.15 K} = 2.56 L$ $V_2 = "9.20 L" ×"37.3 kPa"/"100 kPa" × "273.15 K"/"366.15 K" = "2.56 L"$

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Question #97b37

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