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?

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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?

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Answer 1 · 2015-11-28T03:00:54

${4.838 g H}_{2} O$ $"4.838 g H"_2"O"$ will be formed from the combustion of ${8.064 g C}_{6} H_{12} O_{6}$ $"8.064 g C"_6"H"_12"O"_6"$ .

Explanation:

$C_{6} H_{12} O_{6} {+ 6O}_{2}$ $"C"_6"H"_12"O"_6" + 6O"_2"$ $\to$ $rarr$ ${6CO}_{2} {+ 6H}_{2} O$ $"6CO"_2" + 6H"_2"O"$

Molar Masses of Glucose and Water
$C_{6} H_{12} O_{6} :$ $"C"_6"H"_12"O"_6":$ $180.15588 g/mol$ $"180.15588 g/mol"$
$H_{2} O :$ $"H"_2"O":$ $18.01528 g/mol$ $"18.01528 g/mol"$
https://pubchem.ncbi.nlm.nih.gov

Divide $8.064 g$ $"8.064 g"$ of glucose by its molar mass to get moles glucose.
Multiply times the mol ratio $\frac{6 {mol H}_{2} O}{1 {mol 8C}_{6} H_{12} O_{6}}$ $(6"mol H"_2"O")/(1"mol 8C"_6"H"_12"O"_6")$ from the balanced equation to get moles water.
Multiply times the molar mass of $H_{2} O$ $"H"_2"O"$ to get mass of water.

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

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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?

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Explanation:

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