Question

IV.2. The following problem considers the combustion of butane in a torch. Molecular weight of butane 58.0 g/mol. 2C4H10 + 1302 → 8 CO2 + 10 H2O If 3.00 g of O2 (32.0 g/mol) and 1.00 g of butane are injected into the torch. Under this conditions (10 points each): a) Is there a limiting reagent? if your answer is affirmative, indicate the limiting reactant. b) How many grams of butane are consumed? c) How many grams of CO2 (44.0 g/mol) are produced?

Answer 1

Answer:

(a) Oxygen

(b) 0.84 g

(c) 2.54 g

Explanation:

(a)

Explanation:

The formula for the calculation of moles is shown below:

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

Given: For butane

Given mass = 1.00 g

Molar mass of butane = 58.0 g/mol

Moles of butane = 1.00 g / 58.0 g/mol = 0.0172 moles

Given: For $O_2$

Given mass = 3.00 g

Molar mass of $O_2$ = 32.0 g/mol

Moles of $O_2$ = 3.00 g / 32.0 g/mol = 0.09375 moles

According to the given reaction:

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

2 moles of butane react with 13 moles of $O_2$

1 mole of butane react with 13/2 moles of $O_2$

0.0172 moles  of butane react with (13/2)*0.0172 moles of $O_2$

Moles of $O_2$ required = 0.1118 moles

Available moles of $CuSO_4$ = 0.09375 moles

Limiting reagent is the one which is present in small amount. Thus, $O_2$ is limiting reagent. (0.09375 < 0.1118 )

(b)

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

13 moles of $O_2$ react with  2 moles of butane

1 mole of $O_2$ react with  2/13 moles of butane

0.09375 mole of $O_2$ react with  (2/13)*0.09375  moles of butane

Moles of butane used = 0.0144 moles

Molar mass of butane = 58.0 g/mol

Mass of butane used = Moles × Molar mass = 0.0144 × 58.0 g = 0.84 g

(c)

13 moles of $O_2$ on reaction forms 8 moles of carbon dioxide

1 mole of $O_2$ on reaction forms 8/13 moles of carbon dioxide

0.09375 mole of $O_2$ on reaction forms (8/13)*0.09375 moles of carbon dioxide

Moles of carbon dioxide obtained = 0.05769 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.05769 × 44.0 g = 2.54 g