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 · 2015-11-23T21:30:55

Between 0.5 and 1.0 g of water.

#C_6H_12O_6(s) + 6O_2(g) rarr 6CO_2(g) + 6H_2O(g)#

Moles of glucose, #=# #(8.064*cancel(g))/(180.16*cancel(g)*mol^-1)# #=# #4.476xx10^-2# #mol#.

From the equation above, there will be approximately 0.8 g water produced.

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?