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Question
Calculate the volume of HCl gas formed and chlorine gas required when 40 ml of methane reacts completely with chlorine at S.T.P.
\[\ce{CH4 + 2Cl2 → CH2Cl2 + 2HCl}\]
Solution
\[\ce{CH4 + 2Cl2 → CH2Cl2 + 2HCl}\]
1V 2V 2V
From equation, 1V of CH4 gives = 2 V HCl
so, 40 ml of methane gives = 80 ml of CO2
For 1V of methane = 2V of Cl2 required
So, for 40ml of methane = 40 × 2 = 80 ml of Cl2
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