# What mass of C CL_4 is formed by the reaction of 108.74 g ofmethane with 91.72 g of chlorine?

Jul 29, 2017

Approx. $38 \cdot g$ ${\text{CCl}}_{4}$......

#### Explanation:

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

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

And (ii) equivalent quantities of methane and dichlorine,

$\text{Moles of methane}$ $=$ $\frac{108.74 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} = 6.78 \cdot m o l$

$\text{Moles of dichlorine}$ $=$ $\frac{91.72 \cdot g}{70.90 \cdot g \cdot m o {l}^{-} 1} = 1.29 \cdot m o l$

Clearly chlorine gas is the limiting reagent. And at most we can make $\frac{1.29}{4} \cdot m o l$ $\text{carbon tet} \ldots . = \frac{153.81 \cdot g \cdot m o {l}^{-} 1}{4} = 38.5 \cdot g$ perhalide.