Question

When 20.0 g C2H6 and 60.0 g O2 react to form CO2 and H2O, how many grams of water are formed?

Answer 1

Molar mass of C2H6 = 2*MM(C) + 6*MM(H)

= 2*12.01 + 6*1.008

= 30.068 g/mol

mass of C2H6 = 20.0 g

we have below equation to be used:

number of mol of C2H6,

n = mass of C2H6/molar mass of C2H6

=(20.0 g)/(30.068 g/mol)

= 0.6652 mol

Molar mass of O2 = 32 g/mol

mass of O2 = 60.0 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(60.0 g)/(32 g/mol)

= 1.875 mol

we have the Balanced chemical equation as:

2 C2H6 + 7 O2 ---> 6 H2O + 4 CO2

2 mol of C2H6 reacts with 7 mol of O2

for 0.6652 mol of C2H6, 2.328 mol of O2 is required

But we have 1.875 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 7 mol of O2 reacts, 6 mol of H2O is formed

mol of H2O formed = (6/7)* moles of O2

= (6/7)*1.875

= 1.607 mol

we have below equation to be used:

mass of H2O = number of mol * molar mass

= 1.607*18.02

= 28.95 g

Answer: 29.0 g