Question

Na2O2 + CO2 → Na2CO3 + O2

What volume of O2 gas is produced from 2.80 liters of CO2 at STP?

Answer 1

Answer: The volume of
`O_2` produced will be 1.4 liters.

Step-by-step explanation:

AT STP, the conditions are:

1 mole of a gas occupies 22.4 liters of volume.

The balanced chemical equation for the reaction follows:

`2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2`

By Stoichiometry of the reaction:

2 moles of carbon dioxide produces 1 mole of oxygen gas, which means that

44.8 liters of carbon dioxide gas produces 22.4 liters of oxygen gas.

So, 2.8 liters of carbon dioxide gas will produce =
`(22.4)/(44.8)* 2.8=1.4liters` of oxygen gas.

Hence, the volume of
`O_2` produced will be 1.4 liters.

Answer 2

The volume of O2 gas that is produced from 2.80 L of CO2 at STP is 1.40 L

calculation

write a balanced chemical equation

2Na2O2 + 2CO2 → 2Na2CO3 +O2

by use of mole ratio between CO2:O2 which is 2:1 the number of liters of O2 produced is therefore= 2.8 x 1/2 =1.40 Liters