# Question #ee172

Mar 3, 2017

We need (i) a stoichiometric equation...........

#### Explanation:

$\text{CH"_"4"(g) + "4Cl"_2(g)rarr "CCl"_4 +"4HCl} \left(g\right)$

And (ii) equivalent quantities of methane:

$\text{Moles of methane} = \frac{438.0 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} = 27.3 \cdot m o l$.

Given the 1:1 equivalence, we could make $27.3 \cdot m o l$ of $\text{carbon tetrachloride}$, a mass of.........

$27.3 \cdot m o l \times 153.82 \cdot g \cdot m o {l}^{-} 1 \cong 4.2 \cdot k g$. You will have to convert this into grams.