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
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 Answer: 6.15 g