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The heat liberated when 50 $cm^3$ of 0.2 N $H_2SO_4$ is mixed with 50 $cm^3$ of 1 N KOH, is


(a) 11.46 kJ
(b) 57.3 kJ
(c) 573 kJ
(d) 573 J
Can you answer this question?
 
 

1 Answer

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Answer: 573 J
Here, $H_2SO_4$ is the limiting reactant.
Since, $N_1V_1 = N_2V_2$
$0.2 \times 50 = 1 \times V_2$
$V_2 = \frac{2}{10} \times 50$ = 10 mL
Thus, 50 mL of 0.2 N $H_2SO_4$ will exactly neutralise with 10 mL of 1 N KOH.
Heat produced on neutralisation of 1000 mL of 1 N KOH = 57.3 kJ
$\therefore$ Heat produced on neutralisation of 10 mL of 1 N KOH = $\frac{57.3}{1000}\times 10 \times 1000 $ J = 573 J
answered Feb 25, 2014 by mosymeow_1
 

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