What mass of #"CCl"_4# is required to react completely with #"24.7 g CH"_4#?

The reaction is #"CH"_4+"CCl"_4##rarr##"2CH"_2"Cl"_2#.

1 Answer
Dec 15, 2014

Answer:

#"237 g CCl"_4# is required to react completely with #"24.7 g CH"_4#

Explanation:

Start with a balanced equation.

#"CH"_4+"CCl"_4##rarr##"2CH"_2"Cl"_2#

Convert mass of methane to moles of methane .

Divide the given mass of methane by its molar mass #("16.04 g/mol")#. I prefer to divide by multiplying by the inverse of the molar mass, mol/g.

#24.7color(red)cancel(color(black)("g CH"_4))xx(1"mol CH"_4)/(16.04color(red)cancel(color(black)("g CH"_4)))="1.54 mol CH"_4#

Convert moles #"CH"_4# to moles #"CCl"_4#.

Multiply mol #"CH"_4# by the mol ratio between #"CCl"_4# and #"CH"_4# from the balanced equation, with #"CCl"_4# in the numerator.

#1.54color(red)cancel(color(black)("mol CH"_4))xx(1"mol CCl"_4)/(1color(red)cancel(color(black)("mol CH"_4)))="1.54 mol CCl"_4#

Calculate the mass of #"CCl"_4#.

Multiply mol #"CCl"_4# by its molar mass #("153.82 g/mol")#.

#1.54color(red)cancel(color(black)("mol CCl"_4))xx(153.82"g CCl"_4)/(1color(red)cancel(color(black)("mol CCl"_4)))= "237 g CCl"_4"#

#"237 g CCl"_4# are required to react completely with #"24.7 g CH"_4#