# Question 9f667

Mar 13, 2015

Your reaction needs $\text{203 g}$ of carbon tetrachloride.

$C {H}_{4 \left(g\right)} + {\text{CCl}}_{4 \left(l\right)} \to 2 C {H}_{2} C {l}_{2 \left(g\right)}$

Notice that you have a $\text{1:1}$ mole ratio between methane and carbon tetrachloride, which means that, for every 1 mole of the former, you'll need exactly 1 mole of the latter.

You can determine how many moles of methane you have by using its molar mass

${\text{21.2 g CH"_4 * "1 mole"/"16.04 g" = "1.322 moles CH}}_{4}$

This means that the number of moles of carbon tetrachloride needed is

${\text{1.322 moles CH"_4 * "1 mole CCl"_4/"1 mole CH"_4 = "1.322 moles CCl}}_{4}$

Now use carbon tetrachloride's molar mass to calculate how many grams are needed

$\text{1.322 moles CCl"_4 * "153.82 g"/"1 mole" = "203.35 g}$ ${\text{CCl}}_{4}$

Rounded to three sig figs, the answer will be

m_("CCl"_4) = "203 g"#