# Question #b00ac

Dec 6, 2016

We need $3$ $\text{g}$ of ${\text{Al"_4"C}}_{3}$ to produce $1$ $\text{g}$ of ${\text{CH}}_{4}$

#### Explanation:

${\text{Al"_4"C"_3+12"H"_2"O"rarr 4"Al"("OH")_3 + 3"CH}}_{4}$

We need one gram of methane. So we can work out how many moles we need using the moles' formula

$\text{n"="m"/"M} = \frac{1}{16}$

So we need $\frac{1}{16}$ moles of methane and using the rules of stoichiometry, we know that we need $\frac{1}{48}$ moles of ${\text{Al"_4"C}}_{3}$. So we need to work out the mass of ${\text{Al"_4"C}}_{3}$ that gives us $\frac{1}{48}$ moles.

$\text{m"="nM} = \frac{1}{48} \times 144 = 3$ $\text{g}$

So we need $3$ $\text{g}$ of ${\text{Al"_4"C}}_{3}$ to produce $1$ $\text{g}$ of ${\text{CH}}_{4}$.