Question

A combustion reaction involves 75.0g of C3H8(g) and 85.0g O2 , what mass of H2O is produced?

Write answer to three significant figures.

NO UNITS in answer.

Answer 1

Molar mass of C3H8,

MM = 3*MM(C) + 8*MM(H)

= 3*12.01 + 8*1.008

= 44.094 g/mol

mass(C3H8)= 75.0 g

use:

number of mol of C3H8,

n = mass of C3H8/molar mass of C3H8

=(75 g)/(44.09 g/mol)

= 1.701 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 85.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(85 g)/(32 g/mol)

= 2.656 mol

Balanced chemical equation is:

C3H8 + 5 O2 ---> 4 H2O + 3 CO2

1 mol of C3H8 reacts with 5 mol of O2

for 1.701 mol of C3H8, 8.505 mol of O2 is required

But we have 2.656 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 = (4/5)* moles of O2

= (4/5)*2.656

= 2.125 mol

use:

mass of H2O = number of mol * molar mass

= 2.125*18.02

= 38.28 g

Answer: 38.3