$4 Li (s) + O_{2} (g) ⟶ 2 {Li}_{2} O (s)$
At 25° C, Mi for this reaction is --598.8 kilo joules per mole of ${Li}_{2} O (s)$ formed. What mass of Li should be reacted with excess $O_{2} (g)$ in order to release 150 kJ?

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0.874 g

3.51 g

7.01 g

1.74 g

answer is C.

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Detailed Solution

$4 {Li (s)}^{} + O_{L} (s) \to 2 {Li}_{2} O (s), Δ_{r} H = 2 x - 598.8 KJ$

$\begin{array}{l} 4 mole Li : - 2 \times 598 .8 kJ \\ \frac{4}{2 \times 598 .8} moles Li : - 11 kJ \\ 150 \frac{44}{2 \times 598 .8} {moles Li}^{} : - 150 kJ \end{array}$

$\frac{598 .8}{2} \times \frac{x}{7} = 150 \Rightarrow x = 3 .51 g$

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