Question

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.23 g of butane is mixed with 12. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Molar mass of C4H10,

MM = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass(C4H10)= 5.23 g

use:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(5.23 g)/(58.12 g/mol)

= 8.999*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 12.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(12 g)/(32 g/mol)

= 0.375 mol

Balanced chemical equation is:

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

2 mol of C4H10 reacts with 13 mol of O2

for 8.999*10^-2 mol of C4H10, 0.5849 mol of O2 is required

But we have 0.375 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (8/13)* moles of O2

= (8/13)*0.375

= 0.2308 mol

use:

mass of CO2 = number of mol * molar mass

= 0.2308*44.01

= 10.16 g

Answer: 10. g