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28 gm lithium is mixed with 48 gm $O_{2}$ to react according to the following reaction.
$Li + O_{2} ⟶ {Li}_{2} O$
The mass of ${Li}_{2} O$ formed is :

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15 gm

45 gm

30 gm

60 gm

answer is D.

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Detailed Solution

Here $O_{2}$ is Limiting Reagert
1/2 mole $O_{2}$ form 1 mile ${Li}_{2} O$
i-e log $O_{2}$ form 30 g ${Li}_{2} O$

so,

$\begin{array}{l} 48 O_{2} form \to \frac{30}{16} \times 48 {gL}_{2} O \\ = 60 g L_{2} O \end{array}$

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