Question

Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Supposed 8.14 g of butane is mixed with 41. 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

12.6 g

Step-by-step explanation:

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

For butane:-

Mass of butane = 8.14 g

Molar mass of butane = 58.12 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (8.14\ g)/(58.12\ g/mol)`

`Moles\ of\ butane= 0.14\ mol`

Given: For
`O_2`

Given mass = 41 g

Molar mass of
`O_2` = 31.9988 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (41\ g)/(31.9988\ g/mol)`

`Moles\ of\ O_2 = 1.28\ mol`

According to the given reaction:

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

2 moles of butane react with 13 moles of oxygen

Also,

1 mole of butane react with 6.5 moles of oxygen

So,

0.14 mole of butane react with 6.5*0.14 moles of oxygen

Moles of oxygen = 0.91 moles

Available moles of
`O_2` = 1.28 moles (Extra)

Limiting reagent is the one which is present in small amount. Thus, butane is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

2 moles of butane forms 10 moles of water

Also,

1 mole of butane forms 10 moles of water

So,

0.14 mole of butane forms 5*0.14 mole of water

Moles of water = 0.7 moles

Molar mass of water = 18 g/mol

So,

Mass of water= Moles × Molar mass = 0.7 × 18 g = 12.6 g