Question

Consider the reaction between hydrogen gas and nitrogen gas to form ammonia: 3 H2(g) + N2(g) → 2 NH3(g). What volume of ammonia (in L) could be produced by the reaction of 59.4 liters of hydrogen with 36.7 liters of nitrogen at a constant pressure and temperature?

Answer 1

Answer:

$V_{NH_3}=39.6L$

Explanation:

Hello,

By assuming STP conditions (0°C and 1atm), we first compute the reacting moles of both hydrogen and nitrogen as shown below via the ideal gas equation:

$n_{N_2}=\frac{PV}{RT}=\frac{1atm*36.7L}{0.082 \frac{atm*L}{mol*K}*273.15K} =1.64molN_2\\n_{H_2}=\frac{PV}{RT}=\frac{1atm*59.4L}{0.082 \frac{atm*L}{mol*K}*273.15K} =2.65molH_2$

Next, one identifies the limiting reagent by computing the moles of hydrogen that completely react with 1.64mol of nitrogen as follows:

$n_{H_2}^{reacting}=1.64molN_2*\frac{3molH_2}{1molN_2}=4.92molH_2$

In such a way, as there are just 2.65 available moles of hydrogen one states that we have spare nitrogen and the hydrogen is the limiting reagent, thus, the yielded moles of ammonia are computed as:

$n_{NH_3}=2.65molH_2*\frac{2molNH_3}{3molH_2}=1.767molNH_3$

Finally, one computes the required volume in liters as:

$V_{NH_3}=\frac{nRT}{P}=\frac{1.767mol*0.082 \frac{atm*L}{mol*K} *273.15K}{1atm}\\V_{NH_3}=39.6L$

Best regards.