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$H_{2} O_{2}$ dissociates into $H_{2} O (l) and O_{2} (g)$ . All the $H_{2} O_{2}$ present in 1 L of 3.36 volume $H_{2} O_{2}$ is allowed to dissociate completely and all the released $O_{2}$ is used for the combustion of 1.56 g of diborane $(B_{2} H_{6})$ . If heat of combustion of $B_{2} H_{6}$ is -1900 kJ/mol, then heat released by combustion of 1.56 g of diborane with $O_{2}$ produced by above $H_{2} O_{2}$ solution is –P kJ. Find value of P/95 . Given 1 mole of a gas occupies 22.4 L at STP.

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Detailed Solution

$B_{2} H_{6} + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O mol of B_{2} H_{6} = 0.06$
Mol of $O_{2}$ required $= 0.06 \times 3 = 0.18 mol$
So O2 is limiting reagent so only 0.05 mol of used for combustion
${ΔH}_{I} = - 0.05 \times 1900 = - 95 kJ$

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