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Methylamine, ${CH}_{3} {NH}_{2}$ , reacts with $O_{2}$ to form $O_{2}$ , $N_{2}$ , and $H_{2} O$ . What amount of $O_{2}$ (in moles) is squired to react completely with 1.00mol of ${CH}_{3} {NH}_{2}$ ?

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3.00

2.25

2.50

4.50

answer is A.

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Detailed Solution

In the following reaction we have,

$2 {CH}_{3} {NH}_{2} + \frac{9}{2} O_{2} ⟶ N_{2} + 2 C_{2} + {SH}_{2} O$

$\frac{moles of {CH}_{3} {NH}_{2}}{2} = \frac{mole of O_{2}}{9 / 2}$

mole of $0_{2} \Rightarrow \frac{9}{2} \times \frac{1}{2} \Rightarrow 9 / 4$

$\Rightarrow 2.25 moles$

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