Determine how much carbon dioxide will be obtained when (O2) 360g glucose is burned if the yield is 70%.

1.Let’s find the amount of substance С6Н12О6:

M (C6H12O6) = 180 g / mol.

n = 360 g: 180 g / mol = 2 mol.

Let’s find the quantitative ratios of substances.

C6H12O6 + 6O2 = 6CO2 + 6H2O.

For 1 mol of С6Н12О6 there are 6 mol of СО2.

Substances are in quantitative ratios of 1: 6.

The amount of CО2 is 6 times more than the amount of C6H12O6.

n (CO2) = 6n (C6H12O6) = 2 × 6 = 12 mol.

Find the volume of CO2.

V = n Vn, where Vn is the molar volume of gas equal to 22.4 l / mol.

V = 12 mol × 22.4 L / mol = 268.8 L.

268.8 l – 100% (according to theory),

V l – 70%,

V = (268.8 l × 70%): 100% = 188.16 l.

Answer: 188.16 liters.