Question

N2 + 3H2 --> 2NH3

what volume of hydrogen gas measure at 100kPa and 25 degress celsius, would have reacted to produce 51.10g of ammonia?

Answer 1

We know from the chemical reaction that for every three moles of ammonia we produce, we need 2 moles of hydrogen. The number of moles and mass can be related by the following equation:\[n = {m \over M}\]. The molecular mass of ammonia (M) is 14 + 3 = 17. Therefore, the number of moles in 51.10 g of ammonia is,\[n = {51.10 [g] \over 17 [g/mol]} = 3.006 [mol]\] From the stoichiometric reaction, \[N_{H2} = {2 \over 3} N_{NH3}\] Therefore, the number of moles of H2 needed to produce 3.006 moles of ammonia is 2.004. From the ideal gas law\[PV = nRT \rightarrow V[m^3] = {nRT \over P} = {2.004 [mol]* 8.314 [J/mol-K]*298 [K] \over 100000 [Pa]}\]