Using the following reaction, if 1.75 moles of nitrogen gas are reacted with excess hydrogen gas, how many moles of $N H_{3}$ $NH_3$ will be produced? $N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 -> 2NH_3$ ?

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Answer 1 · 2016-07-08T04:02:46

3.5 moles

Simply write out the mole ratio of the known and unknown components:

$\frac{n (N H_{3})}{n (N_{2})} = \frac{2}{1}$ $(n(NH_3))/(n(N_2))=2/1$

$n (N H_{3}) = 2 n (N_{2}) = 2 \times 1.75 m o l = 3.5 m o l$ $n(NH_3)=2n(N_2)=2xx1.75mol=3.5mol$

Using the following reaction, if 1.75 moles of nitrogen gas are reacted with excess hydrogen gas, how many moles of NH3NH_3 will be produced? N2+3H2→2NH3N_2 + 3H_2 -> 2NH_3?