Question #d0f0e

Dec 27, 2017

$= 19.2 g {H}_{2} O$

Explanation:

1. Find the molar mass of the involved compounds which are obtainable from the periodic table.

${C}_{6} {H}_{12} {O}_{6} = \frac{180 g}{m o l}$

${H}_{2} O = \frac{18 g}{m o l}$

1. Given the mass of the glucose, per convention, convert the $\text{gram}$ to $\text{mole}$; i.e;

$= 32 \cancel{g {C}_{6} {H}_{12} {O}_{6}} \times \frac{1 m o l {C}_{6} {H}_{12} {O}_{6}}{180 \cancel{g {C}_{6} {H}_{12} {O}_{6}}}$

$= 0.1778 m o l {C}_{6} {H}_{12} {O}_{6}$

1. Referring to the given balanced equation for the mole ratio, find the $\eta {H}_{2} O$ through molar conversion; i.e;

$= 0.1778 \cancel{m o l {C}_{6} {H}_{12} {O}_{6}} \times \frac{6 m o l {H}_{2} O}{1 \cancel{m o l {C}_{6} {H}_{12} {O}_{6}}}$

$= 1.0667 m o l {H}_{2} O$

1. Now, find the $m {H}_{2} O$.

$= 1.0667 \cancel{m o l {H}_{2} O} \times \frac{18 g {H}_{2} O}{1 \cancel{m o l {H}_{2} O}}$

$= 19.2 g {H}_{2} O$

Dec 27, 2017

Please see the step process below;

Explanation:

We have the formula;

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

Mole ratio

$1 : 6 \to 6 : 6$

Hence $1$ mole of glucose gives $6$ moles of water

Mass ratio

$180 g : 192 g \to 264 g : 108 g$

It then means that;

$180 g$ will yield $108 g$ of water

$32 g$ will give $x$ of water

$\Rightarrow \frac{180}{32} = \frac{108}{x}$

Cross multiplying..

$\Rightarrow 180 \times x = 108 \times 32$

$\Rightarrow 180 x = 3456$

Divide both sides by $180$

$\Rightarrow \frac{180 x}{180} = \frac{3456}{180}$

$\Rightarrow \frac{\cancel{180} x}{\cancel{180}} = \frac{3456}{180}$

$\Rightarrow x = \frac{3456}{180}$

$\Rightarrow x = 19.2 g$