Question

Gaseous butane CH3CH22CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . What is the theoretical yield of water formed from the reaction of 0.58g of butane and 2.0g of oxygen gas? Be sure your answer has the correct number of significant digits in it.

Answer 1

Molar mass of C4H10 = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass of C4H10 = 0.58 g

we have below equation to be used:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(0.58 g)/(58.12 g/mol)

= 9.979*10^-3 mol

Molar mass of O2 = 32 g/mol

mass of O2 = 2.0 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(2.0 g)/(32 g/mol)

= 6.25*10^-2 mol

we have the Balanced chemical equation as:

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

2 mol of C4H10 reacts with 13 mol of O2

for 9.979*10^-3 mol of C4H10, 6.487*10^-2 mol of O2 is required

But we have 6.25*10^-2 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of H2O = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

From balanced chemical reaction, we see that

when 13 mol of O2 reacts, 10 mol of H2O is formed

mol of H2O formed = (10/13)* moles of O2

= (10/13)*6.25*10^-2

= 4.808*10^-2 mol

we have below equation to be used:

mass of H2O = number of mol * molar mass

= 4.808*10^-2*18.02

= 0.866 g

Answer: 0.87 g