Question

Gaseous ethane (CH,CH,) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO) and gaseous water (HO). Suppose 5.41 g of ethane is mixed with 6.2 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits. x 5 ?

Answer 1

Molar mass of C2H6,

MM = 2*MM(C) + 6*MM(H)

= 2*12.01 + 6*1.008

= 30.068 g/mol

mass(C2H6)= 5.41 g

use:

number of mol of C2H6,

n = mass of C2H6/molar mass of C2H6

=(5.41 g)/(30.07 g/mol)

= 0.1799 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 6.2 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(6.2 g)/(32 g/mol)

= 0.1938 mol

Balanced chemical equation is:

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

2 mol of C2H6 reacts with 7 mol of O2

for 0.1799 mol of C2H6, 0.6297 mol of O2 is required

But we have 0.1938 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 = (6/7)* moles of O2

= (6/7)*0.1938

= 0.1661 mol

use:

mass of H2O = number of mol * molar mass

= 0.1661*18.02

= 3.0 g

Answer: 3.0 g