Question

For the following reaction, 9.90 grams of propane (C3H8) are allowed to react with 20.5 grams of oxygen gas.

propane (C3H8) (g) + oxygen (g) carbon dioxide (g) + water (g)

What is the maximum amount of carbon dioxide that can be formed? grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

1)

Molar mass of C3H8,

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

= 3*12.01 + 8*1.008

= 44.094 g/mol

mass(C3H8)= 9.9 g

number of mol of C3H8,

n = mass of C3H8/molar mass of C3H8

=(9.9 g)/(44.094 g/mol)

= 0.2245 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 20.5 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(20.5 g)/(32 g/mol)

= 0.6406 mol

Balanced chemical equation is:

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

1 mol of C3H8 reacts with 5 mol of O2

for 0.2245 mol of C3H8, 1.1226 mol of O2 is required

But we have 0.6406 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2,

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

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (3/5)* moles of O2

= (3/5)*0.6406

= 0.3844 mol

mass of CO2 = number of mol * molar mass

= 0.3844*44.01

= 16.9 g

Answer: 16.9 g

2)

O2 is limiting reagent

3)

According to balanced equation

mol of C3H8 reacted = (1/5)* moles of O2

= (1/5)*0.6406

= 0.1281 mol

mol of C3H8 remaining = mol initially present - mol reacted

mol of C3H8 remaining = 0.2245 - 0.1281

mol of C3H8 remaining = 0.0964 mol

Molar mass of C3H8,

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

= 3*12.01 + 8*1.008

= 44.094 g/mol

mass of C3H8,

m = number of mol * molar mass

= 9.64*10^-2 mol * 44.094 g/mol

= 4.25 g

Answer: 4.25 g