# Question #de22e

Sep 12, 2017

Work from the known mass and molar ratios of the equation.

#### Explanation:

As a “stoichiometry” question we will assume that all compounds are in the exact amounts, as shown by the equation. Given the molar quantities, we can work backwards from the one given mass quantity to find the rest of the component values.

${O}_{2}$ is 88g, and its molecular weight is 32 g/mol (for the diatomic molecule). So, we have a total of $\frac{88}{32} = 2.75$ moles of ${O}_{2}$, or $5.50$ moles of $O$.

The given stoichiometric equation shows that includes 9 moles of ${O}_{2}$ (3 in the glucose, 6 'free' oxygen):

$6 C {O}_{2} + 6 {H}_{2} O \to {C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2}$

Adjusting that for the given amount of 2.75 moles, each atomic factor must be reduced by $\frac{2.75}{9} = 0.306$

That means in our actual reaction we have $6 C \times 0.306 = 1.83$ moles carbon and $6 {H}_{2} \times 0.306 = 1.83$ moles of ${H}_{2}$.

Converting those to masses we obtain:
$1.83 m o l \times 12 \left(\frac{g}{m o {l}_{C}}\right) = 22 g$ carbon ($C$).
$1.83 m o l \times 2 \left(\frac{g}{m o {l}_{{H}_{2}}}\right) = 3.66 g$ hydrogen (${H}_{2}$).