Question

Calculate the volume of CO2 produced at 37°C and 1.00 atm when 4.78 g of glucose is used up in the reaction.

Answer 1

The glucose combustion equation is:

C6H12O6 +   O2 ---> CO2 + H2O

The equation needs to be balanced as:

C6H12O6 + 6O2 ---> 6CO2 + 6H2O

with stateS:

C6H12O6(s) + 6O2(g) ---> 6CO2(g) + 6H2O(g)

Therefore, there will be only CO2 and H2O in the gas phase

MW of Glucose = 180 g/mol

mol = mass/MW = 4.78/180 = 0.02655 mol of Glucose

1 mol of glucose ---> 6 mol of CO2 and 6 mol of H2O

0.02655 mol --> 0.02655 *6 = 0.1593 mol of CO2 and H2O

We need to claculate the total volume of CO2 (so just use 0.1593)

PV = nRT

T = 37+273 =310

V = nRT/P = (0.1593)(0.082)(310)/(1) = 4.05

V = 4.05 liter of CO2