Question

Gaseous butane (CH₃(CH₂)₂CH₃) will react with gaseous oxygen (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) suppose 4.1 g of butane is mixed with 20.7 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

Solution :

Balanced chemical equation of reaction occuring :

$2CH_{3}CH_{2}CH_{2}CH_{3}(g) + 13O_{2}(g) \rightarrow 8CO_{2}(g) + 10 H_{2}O(g)$

Mass of butane used = 4.1 grams.

Molar mass of butane = 58 g/mol.

As we know,

No. of moles = Mass ÷ Molar mass

Therefore,

No. of moles of butane =

Mass of butane ÷ Molar mass of butane

= 4.1 g ÷ 58 g/mol

= (4.1 ÷ 58)mol

= 0.0707 mol.

Similarly,

Mass of gaseous oxygen used = 20.7 g

Molar mass of gaseous oxygen = 32 g/mol

Therefore,

No. of moles of oxygen =

Mass of oxygen ÷ Molar mass of oxygen

= 20.7 g ÷ 32 g/mol

= (20.7 ÷ 32) mol

= 0.6469 mol.

By this information, we'll try to know limiting reagent for given reaction occuring with given amounts of reactants. Amount of product formed depends upon limiting reagent only.

Let's look at reaction occuring once again,

$2CH_{3}CH_{2}CH_{2}CH_{3}(g) + 13O_{2}(g) \rightarrow 8CO_{2}(g) + 10 H_{2}O(g)$

According to equation,

For complete combustion of 2 mol of butane, 13 mol of oxygen gas is required.

Thus, for complete combustion of 1 mol of butane,

(13 ÷ 2) mol i.e. 6.5 mol of oxygen gas will be required.

Therefore, for complete combustion of 0.0707 mol of butane,

( 6.5 × 0.0707) mol i.e. 0.4596 mol of oxygen will be required.

However, moles of oxygen available is 0.6469 mol.

It means, oxygen is present in excess,

OR butane will be consumed first. Therefore, limiting reagent is butane. Hence, amount of product formed will depend upon amount of butane present for the chemical reaction.

Now, according to equation,

2 mol of butane produces 10 mol of water

Hence, 1 mol of butane will produce 5 mol of water

Therefore, 0.0707 mol of butane will produce,

(5 × 0.0707) mol of water

= 0.3535 mol of water.

Now, molar mass of water = 18 g /mol

As we know,

No. of moles = Mass ÷ Molar mass

Mass = No. of moles × Molar mass

Therefore,

Mass of water =

No. of moles of water × Molar mass of water

Now, putting values, we get,

Mass of water (g) = 0.3535 mol × 18 g/mol

= (0.3535 × 18) grams

= 6.363 grams. ( Significant figures = 3 ) ( Answer )