Question

If 3.0 liters of oxygen gas react with excess carbon monoxide at STP, how many liters of carbon dioxide can be produced under the same conditions?
2CO (g) + O2 (g) yields 2CO2 (g) 1.5 L 3.0 L 4.5 L 6.0 L

Answer 1

N(CO₂)=2n(O₂)

V(CO₂)=2V(O₂)

V(CO₂)=2*3.0=6.0 L

Answer 2

Answer : The volume of carbon dioxide produced, 6.0 L

Solution : Given,

Volume of oxygen = 3.0 L

First we have to calculate the moles of oxygen.

At STP, 1 mole of gas contains 22.4 L volume of gas.

As, 22.4 L volume of oxygen obtained from 1 mole of oxygen gas

As, 3.0 L volume of oxygen obtained from $\frac{3.0}{22.4}=0.1339$ moles of oxygen gas

Now we have to calculate then moles of carbon dioxide.

The balanced chemical reaction will be,

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

From the reaction we conclude that

1 mole of oxygen react to give 2 moles of carbon dioxide

0.1339 mole of oxygen react to give $2\times 0.1339=0.267$ moles of carbon dioxide

The moles of carbon dioxide gas = 0.267 moles

Now we have to calculate the volume of carbon dioxide.

As, 1 mole of carbon dioxide contains 22.4 L volume of carbon dioxide gas

So, 0.267 mole of carbon dioxide contains $22.4\times 0.267=5.98L$ volume of carbon dioxide gas

The volume of carbon dioxide gas = 5.98 L ≈ 6.0 L

Therefore, the volume of carbon dioxide produced, 6.0 L