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

${C}_{6} {H}_{12} {O}_{6} \left(s\right) + 6 {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(g\right)$
Moles of glucose, $=$ $\frac{8.064 \cdot \cancel{g}}{180.16 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $4.476 \times {10}^{-} 2$ $m o l$.