Calculate the volume fo HCL gas formed and chlorine gas required when 40 mL of methane reacts completely with chlorine at STP.
$C H_{4} + 2 C l_{2} \to C H_{2} C l_{2} + 2 H C l$

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Solution

$C H_{4} + 2 C l_{2} \to C H_{2} C l_{2} + 2 H C l$

Here, we see that 1 mole of $C H_{4}$ react with 2 mole of to form 1 mole of $C H_{2} C l_{2}$ and 2 mole HCl.

it means 1 mole of , forms 2 mole of HCl .

so, 1 volume of forms 2volume of HCl.

hence,
40 ml of forms 2 × 40ml HCl

hence, volume of HCl = 80 ml

1 volume of $C H_{4}$ reacts with 2 volume of chlorine.

40 ml of $C H_{4}$ reacts with 80 ml of chlorine.

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