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Equal volumes of equimolar HCl and $H_{2} {SO}_{4}$ are separately neutralised by dilute NaOH solution, then heats liberated are x kCal and y kCal respectively. Which of the following is true

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x = y /2

x = y

x = 2 y

x = y/3

answer is B.

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Detailed Solution

$\frac{HCl}{1 mole} + N a O H \to N a c l + H_{2} O; ∆ H = - x K . C a l$

$\frac{H_{2} {SO}_{4}}{1 mole} + 2 NaOH \to {Na}_{2} {SO}_{4} + 2 H_{2} O; ∆ H = - y$

$v = 2 x$

$x = \frac{y}{2}$

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