How many grams of carbon dioxide are created from the complete
combustion of 21.3 L of butane at STP?*

167.2 g

Explanation:

Known

VC4H10 = 21.3

T = 0.00 C (convert to Kelvin: 273 K)

P = 1.00 atm

Unknown

m = ?g

1. Write the balanced chemical equation

1 C4H10 + 1O2  >  4 CO2 + 5 H2O

2. Find the volume ratio of Carbon Dioxide to Butane

1 C4H10   4 CO2  = 4 volumes CO2 / 1 volume C4H10

3. Multiply by the known volume of n (butane)

21.3 L C4H10  x  4 volumes CO2 / 1 volume C4H10 = 85.2 L C4H10

4. Use ideal gas law

PV = nRT   solve for n  > n = PV/RT

n= (1.00 atm) (85.2 L) / (0.0821 L atm/mol K) (273)  = 3.80 mol CO2

5. Find molar mass of CO2

1 C x 12 + 2 O x 16 = 44.00

6. Multiply the ideal gas law solution (3.80) by molar mass CO2 (44.00)

3.80 mol CO2 x 44.00 g CO2

= 167.2 g CO2  

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.