# Question 97b37

Mar 13, 2015

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

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

At STP

STP is 100 kPa and 273.15 K

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

${P}_{1} = \text{37.3 kPa"; V_1 = "9.20 L"; T_1 = "93.0 °C" = "366.15 K}$
${P}_{2} = \text{100 kPa"; V_2 = "?"; T_2 = "273.15 K}$

V_2 = V_1 × "Pressure factor" × "Temperature factor"#

${V}_{2} = \text{9.20 L" ×"37.3 kPa"/"100 kPa" × "273.15 K"/"366.15 K" = "2.56 L}$