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

Answer:

12.6 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{8.14\ g}{58.12\ g/mol}$

$Moles\ of\ butane= 0.14\ mol$

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{41\ g}{31.9988\ g/mol}$

$Moles\ of\ O_2 = 1.28\ mol$

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