Question

What is the mass of glucose required to produce 44g of CO2 on complete combustion ?

Answer 1

C6H12O6+6O2=>6CO2+6H2O

Molecular mass of Glucose :

=C6H12O6

=(6×C)+(12×H)+(6×O)

=(6×12)+(12×1)+(6×16)

=72+12+96

=180g

Molecular mass of carbon dioxide

=6(CO2)

=6(C+2×0)

=6(12+(2×16))

=6(12+32)

=6(44)

=254g

180g of glucose produces 254g of CO2

1g of CO2=180/254g of glucose

44g of CO2=>(180/254)×44g of glucose

44g of CO2=30g of glucose


30g of glucose is required to produce 44g of CO2 gas.