Question

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

Answer 1

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

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