Question

For the following reaction, 10.5 grams of carbon monoxide are allowed to react with 9.46 grams of oxygen gas.

carbon monoxide(g) + osygen(g) ⟶ carbon dioxide(g)

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

Answer 1

1)

Molar mass of CO,

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

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass(CO)= 10.5 g

use:

number of mol of CO,

n = mass of CO/molar mass of CO

=(10.5 g)/(28.01 g/mol)

= 0.3749 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 9.46 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(9.46 g)/(32 g/mol)

= 0.2956 mol

Balanced chemical equation is:

2 CO + O2 ---> 2 CO2

2 mol of CO reacts with 1 mol of O2

for 0.3749 mol of CO, 0.1874 mol of O2 is required

But we have 0.2956 mol of O2

so, CO is limiting reagent

we will use CO 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 = (2/2)* moles of CO

= (2/2)*0.3749

= 0.3749 mol

use:

mass of CO2 = number of mol * molar mass

= 0.3749*44.01

= 16.5 g

Answer: 16.5 g

2)

CO is limiting reagent

Answer: CO

3)

According to balanced equation

mol of O2 reacted = (1/2)* moles of CO

= (1/2)*0.3749

= 0.1874 mol

mol of O2 remaining = mol initially present - mol reacted

mol of O2 remaining = 0.2956 - 0.1874

mol of O2 remaining = 0.1082 mol

Molar mass of O2 = 32 g/mol

use:

mass of O2,

m = number of mol * molar mass

= 0.1082 mol * 32 g/mol

= 3.462 g

Answer: 3.46 g