Question

Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 3.49 g of butane is mixed with 22. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 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)= 3.49 g

use:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(3.49 g)/(58.12 g/mol)

= 6.005*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 22.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(22 g)/(32 g/mol)

= 0.6875 mol

Balanced chemical equation is:

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

2 mol of C4H10 reacts with 13 mol of O2

for 6.005*10^-2 mol of C4H10, 0.3903 mol of O2 is required

But we have 0.6875 mol of O2

so, C4H10 is limiting reagent

we will use C4H10 in further calculation

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (10/2)* moles of C4H10

= (10/2)*6.005*10^-2

= 0.3002 mol

use:

mass of H2O = number of mol * molar mass

= 0.3002*18.02

= 5.409 g

Answer: 5.41 g