Question

Gaseous ethane (CH,CH,) will react with gaseous oxygen (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). Suppose 4.21 g of

ethane is mixed with 31. 9 of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has
the correct number of significant digits.

Answer 1

Answer: The mass of
`CO_2` produced is 12.32 g

Step-by-step explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

• For ethane:

Given mass of ethane = 4.21 g

Molar mass of ethane = 30 g/mol

Putting values in equation 1, we get:

`\text{Moles of ethane}=(4.21g)/(30g/mol)=0.140mol`

• For oxygen gas:

Given mass of oxygen gas = 31.9 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(31.9g)/(32g/mol)=0.997mol`

The chemical equation for the combustion of ethane follows:

`2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O`

By stoichiometry of the reaction:

If 2 moles of ethane reacts with 7 moles of oxygen gas

So, 0.140 moles of ethane will react with =
`(7)/(2)* 0.140=0.49mol` of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

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

By the stoichiometry of the reaction:

If 2 moles of ethane produces 4 moles of
`CO_2`

So, 0.140 moles of ethane will produce =
`(4)/(2)* 0.140=0.28mol` of
`CO_2`

We know, molar mass of
`CO_2` = 44 g/mol

Putting values in above equation, we get:

`\text{Mass of }CO_2=(0.28mol* 44g/mol)=12.32g`

Hence, the mass of
`CO_2` produced is 12.32 g