Question

For the following reaction, 8.90 grams of benzene (C6H6) are allowed to react with 12.6 grams of oxygen gas.

benzene (C6H6) (l) + 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 C6H6,

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

= 6*12.01 + 6*1.008

= 78.108 g/mol

mass(C6H6)= 8.9 g

number of mol of C6H6,

n = mass of C6H6/molar mass of C6H6

=(8.9 g)/(78.108 g/mol)

= 0.1139 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 12.6 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(12.6 g)/(32 g/mol)

= 0.3937 mol

Balanced chemical equation is:

2 C6H6 + 15 O2 ---> 12 CO2 + 6 H2O

2 mol of C6H6 reacts with 15 mol of O2

for 0.1139 mol of C6H6, 0.8546 mol of O2 is required

But we have 0.3937 mol of O2

so,

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 = (12/15)* moles of O2

= (12/15)*0.3937

= 0.315 mol

mass of CO2 = number of mol * molar mass

= 0.315*44.01

= 13.86 g

Answer: 13.9 g

2)

O2 is limiting reagent

3)

According to balanced equation

mol of C6H6 reacted = (2/15)* moles of O2

= (2/15)*0.3937

= 0.0525 mol

mol of C6H6 remaining = mol initially present - mol reacted

mol of C6H6 remaining = 0.1139 - 0.0525

mol of C6H6 remaining = 0.0614 mol

Molar mass of C6H6,

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

= 6*12.01 + 6*1.008

= 78.108 g/mol

mass of C6H6,

m = number of mol * molar mass

= 6.144*10^-2 mol * 78.108 g/mol

= 4.80 g

Answer: 4.80 g