Question

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 13. g of butane is mixed with 70.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: The maximum amount of water that could be produced by the chemical reaction is 20.16 grams

Step-by-step explanation:

To calculate the number of moles, we use the equation:

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

• For butane:

Given mass of butane = 13 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

`\text{Moles of butane}=(13g)/(58.12g/mol)=0.224mol`

• For oxygen gas:

Given mass of oxygen gas = 70.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(70.9g)/(32g/mol)=2.216mol`

The chemical equation for the reaction of butane and oxygen gas follows:

`2C_4H_(10)+13O_2\rightarrow 8CO_2+10H_2O`

By Stoichiometry of the reaction:

2 moles of butane reacts with 13 moles of oxygen gas

So, 0.224 moles of butane will react with =
`(13)/(2)* 0.224=1.456mol` of oxygen gas

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

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

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water

So, 0.224 moles of butane will produce =
`(10)/(2)* 0.224=1.12moles` of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.12 moles

Putting values in equation 1, we get:

`1.12mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.12mol* 18g/mol)=20.16g`

Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams