Dichloromethane $(C H_{2} C l_{2})$ , used as a solvent is prepared by reaction of methane $(C H_{4})$ with chlorine $(C l_{2})$ . How many grams of $C H_{2} C l_{2}$ result from reaction of $1.85$ kg of methane if the yield is $43.1 %$ ?

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Solution

The reaction is as follows:
$C H_{4} + C l_{2} \to C H_{2} C l_{2} + 2 H C l$
$1$ mole of methane gives $1$ mole of dichloromethane.
Hence, $1.85$ kg of $43.1 %$ pure methane will give $\frac{43.1}{100} \times \frac{1.85 \times 1000}{16} \times 84.93 = 4236$ g of dichloromethane.

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Dichloromethane (CH2Cl2), used as a solvent is prepared by reaction of methane (CH4) with chlorine (Cl2). How many grams of CH2Cl2 result from reaction of 1.85 kg of methane if the yield is 43.1%?

The reaction is as follows:CH4+Cl2→CH2Cl2+2HCl1 mole of methane gives 1 mole of dichloromethane. Hence, 1.85 kg of 43.1% pure methane will give 43.1100×1.85×100016×84.93=4236 g of dichloromethane.

Dichloromethane $(C H_{2} C l_{2})$ , used as a solvent is prepared by reaction of methane $(C H_{4})$ with chlorine $(C l_{2})$ . How many grams of $C H_{2} C l_{2}$ result from reaction of $1.85$ kg of methane if the yield is $43.1 %$ ?

The reaction is as follows:
$C H_{4} + C l_{2} \to C H_{2} C l_{2} + 2 H C l$
$1$ mole of methane gives $1$ mole of dichloromethane.
Hence, $1.85$ kg of $43.1 %$ pure methane will give $\frac{43.1}{100} \times \frac{1.85 \times 1000}{16} \times 84.93 = 4236$ g of dichloromethane.