Question

The combustion of a sample of butane, C4H10, produced 2.46 grams of water.

Answer 1

Answer:

(A) 0.14 moles water formed.

(B) 0.03 moles butane burned

(C) 5.7 grams of butane burned

(D) 0.18 moles of oxygen was used up in the reaction.

Explanation:

(A)

The given chemical reaction is as follows.

The number of water molecules in the reaction = 10

$2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O$

$\frac{2.46g}{10(18)}=\frac{xH_{2}O}{10H_{2}O}= 0.14$

Therefore, 0.14 moles water formed.

(B)

Molarmass of butane = 180 g

The number of butane molecules in the reaction = 2

$\frac{2.46g}{180}=\frac{xC_{4}H_{10}}{2C_{4}H_{10}}= 0.03$

Therefore, 0.03 moles butane burned

(C)

Molar mass of oxygen = 32g/mol

The number of water molecules in the reaction = 10

The number of oxygen molecules in the reaction = 13

$\frac{2.46g}{10(18)}=\frac{xO_{2}}{13(32)}= 5.7g$

Therefore, 5.7 grams of butane burned

(D)

Molar mass of oxygen = 32 g/mol

Used amount of oxygen in the reaction = 5.7 g

$\frac{5.7}{32}=0.18$

Therefore, 0.18 moles of oxygen was used up in moles.