Question

Read the given equation. Na2O2 + CO2 → Na2CO3 + O2 What volume of O2 gas is produced from 2.80 liters of CO2 at STP? 5.60 liters 4.20 liters 2.10 liters 1.40 liters

Answer 1

The equation should be balanced as follows  2Na2O2 + 2CO2⇒ 2Na2CO3 +O2

2 moles of carbon (IV) oxide give 1 mole of oxygen.

Therefore, the mole ratio is of CO2 to O2 is 2:1

if 2.8 liters of CO2 were used, then the volume of O2 produced will be given by:   (2.80liters× 1)/2= 1.40 liters

Answer 2

Answer:

The correct answer is 1.40 L.

Explanation:

Volume of $CO_2$ = 2.80 L

At STP. 1 mol of gas occupies = 22.4 L

Then 2.80 L of volume will be occupied by:

$\frac{1}{22.4 L}\times 2.80 L=0.125 mol$

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

Acording to reaction 2 mol of $CO_2$ gives 1mol of $O_2$

Then 0.125 mol of $CO_2$ will give:

$\frac{1}{2}\times 0.125 mol=0.0625 mol$ of $O_2$

Volume occupied by 0.0625 mol of oxygen gas at STP :

$0.0625 \times 22.4 L= 1.40 L$

1.40 L of oxygen gas will be produced.