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:
The correct answer is 1.40 L.
Explanation:
Volume of [tex]CO_2[/tex] = 2.80 L
At STP. 1 mol of gas occupies = 22.4 L
Then 2.80 L of volume will be occupied by:
[tex]\frac{1}{22.4 L}\times 2.80 L=0.125 mol[/tex]
[tex]2Na_2O_2 + 2CO_2 \rightarrow 2Na_2CO_3 + O_2[/tex]
Acording to reaction 2 mol of [tex]CO_2[/tex] gives 1mol of [tex]O_2[/tex]
Then 0.125 mol of [tex]CO_2[/tex] will give:
[tex]\frac{1}{2}\times 0.125 mol=0.0625 mol [/tex] of [tex]O_2[/tex]
Volume occupied by 0.0625 mol of oxygen gas at STP :
[tex] 0.0625 \times 22.4 L= 1.40 L[/tex]
1.40 L of oxygen gas will be produced.
