Question

ex. a) how many moles of ammonia are produced when 9.0 mol of hydrogen gas react with 5.0 mol of nitrogen gas?
b) how many moles of excess reactant remain after the reaction is complete?

Explanation:

Step1: 写出化学反应方程式并确定比例关系

化学反应方程式为\(N_{2}+3H_{2}\to2NH_{3}\)，由方程式可知\(N_{2}\)与\(H_{2}\)反应的物质的量之比为\(1:3\)。

Step2: 判断过量反应物

已知\(n(H_{2}) = 9.0mol\)，\(n(N_{2}) = 5.0mol\)。若\(N_{2}\)完全反应，需要\(H_{2}\)的物质的量\(n_{1}(H_{2})=5.0mol\times3 = 15.0mol\)，但实际\(H_{2}\)只有\(9.0mol\)，所以\(H_{2}\)是不足量的，\(N_{2}\)是过量的。

Step3: 计算反应消耗的\(N_{2}\)的物质的量

根据\(N_{2}\)与\(H_{2}\)反应的物质的量之比\(1:3\)，设反应消耗\(N_{2}\)的物质的量为\(x\)，则\(\frac{x}{9.0mol}=\frac{1}{3}\)，解得\(x = 3.0mol\)。

Step4: 计算剩余过量反应物\(N_{2}\)的物质的量

剩余\(N_{2}\)的物质的量\(n_{剩余}(N_{2})=5.0mol - 3.0mol=2.0mol\)。

Answer:

\(2.0mol\)