Question

Acetylene (ethyne) gas reacts with oxygen and burns at ${3300}^{o}C$in an acetylene torch. role="math" localid="1652782652190" $(7.4,7.7,8.6,11.4,11.5)$

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 $8.5L$of acetylene gas at STP?

d. How many liters of $C{O}_2$ gas at STP are produced when $30.0g$ of acetylene undergoes combustion?

Short answer

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

$2H_2{C}_2+5O_2\to 4C{O}_2+2H_{2}O$

(b) The molar mass of acetylene is $26.03g/mol$.

(c) $30.35g$of $O_2$are needed to react with $8.5L$of acetylene gas at STP.

(d) $51.69LC{O}_2$gas at STP are produced when$30.0g$of acetylene undergoes combustion.

Step-by-step solution

Part(a) Step 1: Given information

We have been given,

The balanced chemical equation for the complete combustion of acetylene.

Part(a) Step 2: Explanation

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

$2H_2{C}_2+5O_2\to 4C{O}_2+2H_{2}O$.

Part(b) Step 1: Given information

We have to find out the molar mass of acetylene.

Part(b) Step 2: Explanation

As we know the formula of acetylene is $C_2{H}_2$,

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

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

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

Molecular mass of acetylenerole="math" localid="1652784231246" $=12\times 2+1\times 2=26g/mol.$

Part(c) Step 1: Given information

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

Part(c) Step 2: Explanation

Now, we solve this part to get answer,

$C_2{H}_2+\frac{5}{2}{O}_2\to 2C{O}_2+H_{2}O$,

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

$1mole{C}_2{H}_2=22.4L$,

And $1mole$$O_2$$=32g$,

$\Rightarrow \frac{5}{2}mole{O}_2=\frac{5}{2}\times 32g,$

$\Rightarrow 80g$,

$\Rightarrow 22.4Lof{C}_2{H}_2$reacts with $80g$of $O_2$,

$1L$of $C_2{H}_2$reacts with $\frac{80}{22.4}g$of $O_2$,

$\therefore$$8.5L$of $C_2{H}_2$reacts with $\frac{80\times 8.5}{22.4}$of $O_2$,

$\Rightarrow 30.35g$.

Part(d) Step 1:Given information

We have to find out the how many liters of $C{O}_2$gas at STP are produced when $30.0g$of acetylene undergoes combustion.

Part(d) Step 2: Explanation

Now, we solve this question to get answer,

$C_2{H}_2+\frac{5}{2}{O}_2\to 2C{O}_2+H_2$,

we know,

$44gC{O}_2=22.4L$,

Hence, $2\times 44gC{O}_2=\frac{22.4\times 2\times 44}{44}L$,

$\Rightarrow 44.8L$

$26g$of $C_2{H}_2$reacts with $44.8L$of $C{O}_2$,

$1g$of $C_2{H}_2$reacts with $\frac{44.8}{26}L$of $C{O}_2$,

$\therefore 30g$of $C_2{H}_2$reacts with $\frac{44.8\times 30}{26}L$of $C{O}_2$,

$\Rightarrow 51.69L$of$C{O}_2$.