Question

Question #d0f0e

Answer 1

`=19.2gH_2O`

Explanation:

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

`C_6H_12O_6=(180g)/(mol)`

`H_2O=(18g)/(mol)`

1. Given the mass of the glucose, per convention, convert the `"gram"` to `"mole"`; i.e;

`=32cancel(gC_6H_12O_6)xx(1molC_6H_12O_6)/(180cancel(gC_6H_12O_6))`

`=0.1778molC_6H_12O_6`

1. Referring to the given balanced equation for the mole ratio, find the `etaH_2O` through molar conversion; i.e;

`=0.1778cancel(molC_6H_12O_6)xx(6molH_2O)/(1cancel(molC_6H_12O_6))`

`=1.0667molH_2O`

1. Now, find the `mH_2O`.

`=1.0667cancel(molH_2O)xx(18gH_2O)/(1cancel(molH_2O))`

`=19.2gH_2O`

Answer 2

Please see the step process below;

Explanation:

We have the formula;

`C_6H_12O_6 + 6O_2 -> 6CO_2 + 6H_2O`

Mole ratio

`1 : 6 -> 6 : 6`

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

Mass ratio

`180g : 192g -> 264g : 108g`

It then means that;

`180g` will yield `108g` of water

`32g` will give `x` of water

`rArr 180/32 = 108/x`

Cross multiplying..

`rArr 180 xx x = 108 xx 32`

`rArr 180x = 3456`

Divide both sides by `180`

`rArr (180x)/180 = 3456/180`

`rArr (cancel180x)/cancel180 = 3456/180`

`rArr x = 3456/180`

`rArr x = 19.2g`