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For a sequential reaction,
$\begin{aligned} {NH}_{3} ⟶ N_{2} + H_{2} . . . (i) \\ H_{2} + O_{2} ⟶ H_{2} O . . . . (ii) \end{aligned}$
What will be the amount of water which will be obtained if 5 moles of ${NH}_{3}$ is mixed with 3 moles of $O_{2}$ and % yield of 1st and 2nd reaction is 50% and 80% respectively?

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3 moles

2 moles

2.5 mole

2.4 moles

answer is A.

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Detailed Solution

${NH}_{3} \overset{{sO %}_{3}}{⟶} \frac{1}{2} N_{2} + \frac{3}{2} H_{2}$

$H_{2} + \frac{1}{2} O_{2} \overset{80 %}{⟶} H_{2} O$

$(LR)$

$n_{H_{2}}$ formed $= \frac{5}{2} \times \frac{3}{2}$

$n_{H_{2} O}$ formed $= \frac{15}{4} \times 0.8 = 3$ moles.

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