# The equation for the combustion of glucose is: C6H12O6(s) + 6O2(g) -->6CO2(g) + 6H2O(g). How many grams of H2O will be produced when 8.064g of glucose is burned?

Nov 28, 2015

$\text{4.838 g H"_2"O}$ will be formed from the combustion of $\text{8.064 g C"_6"H"_12"O"_6}$.

#### Explanation:

$\text{C"_6"H"_12"O"_6" + 6O"_2}$$\rightarrow$$\text{6CO"_2" + 6H"_2"O}$

Molar Masses of Glucose and Water
$\text{C"_6"H"_12"O"_6} :$$\text{180.15588 g/mol}$
$\text{H"_2"O} :$$\text{18.01528 g/mol}$
https://pubchem.ncbi.nlm.nih.gov

1. Divide $\text{8.064 g}$ of glucose by its molar mass to get moles glucose.
2. Multiply times the mol ratio $\left(6 \text{mol H"_2"O")/(1"mol 8C"_6"H"_12"O"_6}\right)$ from the balanced equation to get moles water.
3. Multiply times the molar mass of $\text{H"_2"O}$ to get mass of water.

8.064cancel("g C"_6"H"_12"O"_6)xx(1cancel("mol C"_6"H"_12"O"_6))/(180.15588cancel("g C"_6"H"_12"O"_6))xx(6cancel("mol H"_2"O"))/(1cancel("mol C"_6"H"_12"O"_6))xx(18.01528"g H"_2"O")/(1cancel("mol H"_2"O"))="4.838 g H"_2"O"