Question #34863

For the reaction
?CH4+?Cl2 !?CCl4+?HCl ,
what is the maximum amount of CCl4
(153.823 g/mol) which could be formed from
13.79 g of CH4 (16.0425 g/mol) and 11.41 g
of Cl2 (70.906 g/mol)?
Answer in units of g

Expert's answer

First of all it is necessary to balance the chemical equation

CH_{4} + 4Cl_{2} = CCl_{4} + 4HCl

the amount of substance of CH_{4} is calculated by formula:

n = m/M

n(CH_{4}) = 13.79 g/16.0425 g/mol = 0.86 mol

the amount of substanse of Cl_{2} is calculated by the same formula

n(Cl_{2}) = 11.41g/70.906 g/mol = 0.16 mol

the ratio among the amounts is

n(CH_{4}) : n(Cl_{2}) = 1 : 4

Therefore, the methane is in excess.

Hence the amount of substance of CCl_{4} is

n(Cl_{2}) : n(CCl_{4}) = 4:1

or

0.16 : n(CCl_{4}) = 4:1

n(CCl_{4}) = 0.04 mol

m(CCl_{4}) = 0.04 mol * 153.823 g/ mol = 6.15 g

Answer: 6.15 g

CH

the amount of substance of CH

n = m/M

n(CH

the amount of substanse of Cl

n(Cl

the ratio among the amounts is

n(CH

Therefore, the methane is in excess.

Hence the amount of substance of CCl

n(Cl

or

0.16 : n(CCl

n(CCl

m(CCl

Answer: 6.15 g

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