Question

Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 5.23 g of butane is mixed with 7.7 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 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

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)= 7.7 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(7.7 g)/(32 g/mol)

= 0.2406 mol

Balanced chemical equation is:

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

2 mol of C4H10 reacts with 13 mol of O2

for 0.089986 mol of C4H10, 0.584911 mol of O2 is required

But we have 0.240625 mol of O2

so, O2 is limiting reagent

we will use O2 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 = (10/13)* moles of O2

= (10/13)*0.240625

= 0.185096 mol

mass of H2O = number of mol * molar mass

= 0.1851*18.02

= 3.335 g

Answer: 3.3 g