Question

the complete combustion of 600mL of ethane used 2500mL of neutral gas. Calculate the volume of unused neutral gas and volume of acidic gas formed.

Answer 1

Answer: The volume of unused neutral gas is 1000 mL and volume of acidic gas is 1200 mL

Explanation:

At STP:

1 mole of a gas occupies 22.4 L or 22400 mL of volume

We are given:

Volume of ethane = 600 mL

Volume of neutral gas (oxygen gas) = 2500 mL

The chemical equation for the combustion of ethane follows:

$2C_2H_6+5O_2\rightarrow 4CO_2+6H_2O$

Here, the neutral gas is oxygen gas and acidic gas is carbon dioxide gas

By Stoichiometry of the reaction:

$(2\times 22400)mL$ of ethane reacts with $(5\times 22400)mL$ of oxygen gas

So, 600 mL of ethane will react with = $\frac{(5\times 22400)}{(2\times 22400)}\times 600=1500mL$ of oxygen gas

As, the given amount of oxygen gas is less than the required amount. So, it is considered as an excess reagent

Then, ethane is considered as a limiting reagent because it limits the formation of product.

• Volume of oxygen (neutral) gas left = 2500 - 1500 = 1000 mL

By Stoichiometry of the reaction:

$(2\times 22400)mL$ of ethane produces tex](4\times 22400)mL[/tex] of carbon dioxide gas

So, 600 mL of ethane will produce $\frac{(4\times 22400)}{(2\times 22400)}\times 600=1200mL$ of carbon dioxide gas

Volume of acidic gas (carbon dioxide) produced = 1200 mL

Hence, the volume of unused neutral gas is 1000 mL and volume of acidic gas is 1200 mL