Q. 11.68

Question

Acetylene (ethyne) gas reacts with oxygen and burns at $3300^{o} C$ in an acetylene torch. $(7.4, 7.7, 8.6, 11.4, 11.5)$

a. Write the balanced chemical equation for the complete combustion of acetylene.

b. What is the molar mass of acetylene?

c. How many grams of oxygen are needed to react with $8.5 L$ of acetylene gas at STP?

d. How many liters of $C O_{2}$ gas at STP are produced when $30.0 g$ of acetylene undergoes combustion?

Step-by-Step Solution

Verified

Answer

(a) The balanced chemical equation for the complete combustion of acetylene is

$2 H_{2} C_{2} + 5 O_{2} \to 4 C O_{2} + 2 H_{2} O$

(b) The molar mass of acetylene is $26.03 g / m o l$ .

(d) $51.69 L C O_{2}$ gas at STP are produced when $30.0 g$ of acetylene undergoes combustion.

1Part(a) Step 1: Given information

We have been given,

The balanced chemical equation for the complete combustion of acetylene.

2Part(a) Step 2: Explanation

Now, we write the balanced chemical equation for the complete combustion of acetylene.

$2 H_{2} C_{2} + 5 O_{2} \to 4 C O_{2} + 2 H_{2} O$ .

3Part(b) Step 1: Given information

We have to find out the molar mass of acetylene.

4Part(b) Step 2: Explanation

As we know the formula of acetylene is $C_{2} H_{2}$ ,

Mass of carbon (C) $= 12 g$ ,

Mass of the Hydrogen(H) $= 1 g$ ,

Mass of Oxygen(O) $= 16 g$ ,

Molecular mass of acetylene $= 12 \times 2 + 1 \times 2 = 26 g / m o l .$

5Part(c) Step 1: Given information

We have to find out the how many grams of oxygen are needed to react with $8.5 L$ of acetylene gas at STP.

6Part(c) Step 2: Explanation

Now, we solve this part to get answer,

$C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 C O_{2} + H_{2} O$ ,

$1$ mole $\frac{5}{2}$ mole,

$1 m o l e C_{2} H_{2} = 22.4 L$ ,

And $1 m o l e$ $O_{2}$ $= 32 g$ ,

$\Rightarrow \frac{5}{2} m o l e O_{2} = \frac{5}{2} \times 32 g,$

$\Rightarrow 80 g$ ,

$\Rightarrow 22.4 L o f C_{2} H_{2}$ reacts with $80 g$ of $O_{2}$ ,

$1 L$ of $C_{2} H_{2}$ reacts with $\frac{80}{22.4} g$ of $O_{2}$ ,

$∴$ $8.5 L$ of $C_{2} H_{2}$ reacts with $\frac{80 \times 8.5}{22.4}$ of $O_{2}$ ,

$\Rightarrow 30.35 g$ .

7Part(d) Step 1:Given information

We have to find out the how many liters of $C O_{2}$ gas at STP are produced when $30.0 g$ of acetylene undergoes combustion.

8Part(d) Step 2: Explanation

Now, we solve this question to get answer,

$C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 C O_{2} + H_{2}$ ,

we know,

$44 g C O_{2} = 22.4 L$ ,

Hence, $2 \times 44 g C O_{2} = \frac{22.4 \times 2 \times 44}{44} L$ ,

$\Rightarrow 44.8 L$

$26 g$ of $C_{2} H_{2}$ reacts with $44.8 L$ of $C O_{2}$ ,

$1 g$ of $C_{2} H_{2}$ reacts with $\frac{44.8}{26} L$ of $C O_{2}$ ,

$∴ 30 g$ of $C_{2} H_{2}$ reacts with $\frac{44.8 \times 30}{26} L$ of $C O_{2}$ ,

$\Rightarrow 51.69 L$ of $C O_{2}$ .

Q. 11.65

Q. 11.57

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Q. 11.56

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