Question

How many grams of carbon dioxide are created from the complete

combustion of 21.3 L of butane at STP?*​

Answer 1

167.2 g

Step-by-step 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