Question

3. Consider this reaction: 2 C2H6 + 702 4CO2 + 6H2O (a) How many grams of oxygen are required to completely react with (burn) 100.0 g of ethane?

Answer 1

a)

Molar mass of C2H5,

MM = 2*MM(C) + 5*MM(H)

= 2*12.01 + 5*1.008

= 29.06 g/mol

mass of C2H5 = 1*10^2 g

mol of C2H5 = (mass)/(molar mass)

= 1*10^2/29.06

= 3.441 mol

According to balanced equation

mol of O2 required = (7/2)* moles of C2H5

= (7/2)*3.441

= 12.04 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 12.04*32

= 385 g

Answer: 385 g

b)

From part 1, we can say that for given mass of C2H6, more mass of O2 is required.

Here mass of O2 and C2H6 are same.

So, O2 is limiting reagent.

We will us eO2 in further calculations.

Molar mass of O2 = 32 g/mol

mass of O2 = 55 g

mol of O2 = (mass)/(molar mass)

= 55/32

= 1.719 mol

According to balanced equation

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

= (6/7)*1.719

= 1.473 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = number of mol * molar mass

= 1.473*18.02

= 26.54 g

Answer: 26.5 g