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# Answer to Question #34863 in Other Chemistry for angela

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)?
First of all it is necessary to balance the chemical equation
CH4 + 4Cl2 = CCl4 + 4HCl
the amount of substance of CH4 is calculated by formula:
n = m/M
n(CH4) = 13.79 g/16.0425 g/mol = 0.86 mol
the amount of substanse of Cl2 is calculated by the same formula
n(Cl2) = 11.41g/70.906 g/mol = 0.16 mol
the ratio among the amounts is
n(CH4) : n(Cl2) = 1 : 4
Therefore, the methane is in excess.
Hence the amount of substance of CCl4 is
n(Cl2) : n(CCl4) = 4:1
or
0.16 : n(CCl4) = 4:1
n(CCl4) = 0.04 mol
m(CCl4) = 0.04 mol * 153.823 g/ mol = 6.15 g

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