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

## The reaction is ${\text{CH"_4+"CCl}}_{4}$$\rightarrow$${\text{2CH"_2"Cl}}_{2}$.

Dec 15, 2014

${\text{237 g CCl}}_{4}$ is required to react completely with ${\text{24.7 g CH}}_{4}$

#### Explanation:

${\text{CH"_4+"CCl}}_{4}$$\rightarrow$${\text{2CH"_2"Cl}}_{2}$

Convert mass of methane to moles of methane .

Divide the given mass of methane by its molar mass $\left(\text{16.04 g/mol}\right)$. 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 ${\text{CH}}_{4}$ to moles ${\text{CCl}}_{4}$.

Multiply mol ${\text{CH}}_{4}$ by the mol ratio between ${\text{CCl}}_{4}$ and ${\text{CH}}_{4}$ from the balanced equation, with ${\text{CCl}}_{4}$ 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 ${\text{CCl}}_{4}$.

Multiply mol ${\text{CCl}}_{4}$ by its molar mass $\left(\text{153.82 g/mol}\right)$.

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"

${\text{237 g CCl}}_{4}$ are required to react completely with ${\text{24.7 g CH}}_{4}$