# How do you use stoichiometry to convert moles to mass?

Aug 4, 2018

Suppose we completely combusted an $8 \cdot g$ mass of methane...

#### Explanation:

Clearly, we could write the stoichiometric reaction, the which gives us the absolute molar equivalence with respect to product and reactants...

${\underbrace{C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) + \Delta}}_{\text{as written 80 g of reactant gives 80 g of product}}$

But we proposed an $8 \cdot g$ mass with respect to methane, a molar quantity of $\frac{8 \cdot g}{16.01 \cdot g \cdot m o {l}^{-} 1} = \frac{1}{2} \cdot m o l$...

Given the stoichiometry, therefore we require a $1 \cdot m o l$ quantity of dioxygen gas, i.e. $32 \cdot g$, with which the methane reacts to give a $\frac{1}{2} \cdot m o l$ quantity $C {O}_{2}$, i.e. $22 \cdot g$, and ONE mole of water, i.e. $18 \cdot g$...and thus $40 \cdot g$ of reactant gives PRECISELY $40 \cdot g$ of product....PLUS ENERGY.