Question

What amount of the excess reagent remains when 0.30 mol NH3 reacts with 0.40 mol O2 to produce NO and H2O? 4NH3 +502 + 4NO + 6H20
(A) 0.10 mol NH,
(B) 0.10 mol
(C) 0.025 mol NH
(D) 0.025 mol O2

Answer 1

Answer: (D) 0.025 mol $O_2$

Explanation:

The given balanced equation is :

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

According to stoichiometry :

4 moles of ammonia $(NH_3)$ reacts with = 5 moles of oxygen $(O_2)$

Thus 0.30 moles of  ammonia  $(NH_3)$  reacts with = $\frac{5}{4}\times 0.30=0.375$ moles of oxygen $(O_2)$

Now as given moles of oxygen are more than the required amount, oxygen is the excess reagent.

moles of oxygen $(O_2)$ left = 0.40 - 0.375 = 0.025

Thus 0.025 mol $O_2$ excess reagent remains.