Question

In an acetylene torch, acetylene gas burns in oxygen to produce carbon dioxide, water, and energy.

$2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{∆}{\to }4{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

a. How many moles of O₂ are needed to react with 2.40 moles of C₂H₂?

b. How many moles of CO₂ are produced when 3.5 moles of C₂H₂ react?

c. How many moles of C₂H₂ are needed to produce 0.50 moles of H₂O?

d. How many moles of CO₂ are produced from 0.200 moles of O₂?

Short answer

a.6moles of O₂ are needed to react with 2.40 moles of C₂H₂ .

b.7moles of CO₂ are produced when 3.5 moles of C₂H₂ react .

c. 0.50 moles of C₂H₂ are needed to produce 0.50 moles of H₂O .

d.0.16moles of CO₂ are produced from 0.200 moles of O₂ .

Step-by-step solution

Part (a) Given Information

The moles of C₂H₂ are 2.40 moles.

Part (a) Explanation

The balanced chemical reaction when acetylene burns with air is as follows:

$2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{∆}{\to }4{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

From the above reaction, it is observed that 2 moles of acetylene react with five moles of oxygen, to produce four moles of carbon dioxide and two moles of water.

Therefore, to calculate the number of moles of oxygen, the following conversion factor is required.

$\frac{5\mathrm{moles}{\mathrm{O}}_2}{2\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}\mathrm{and}\frac{2\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}{5\mathrm{moles}{\mathrm{O}}_2}$

So, the number of moles of oxygen required to react with2.40 moles of C₂H₂ is given as follows:

$2.40\cancel{\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}\times \frac{5\mathrm{moles}{\mathrm{O}}_2}{2\cancel{\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}}=2.40\times 2.5\mathrm{moles}{\mathrm{O}}_2=6\mathrm{moles}{\mathrm{O}}_2$

Hence, 6moles of O₂ are needed to react with 2.40 moles of C₂H₂.

Part (a) Given Information

The moles ofC₂H₂are 3.5 moles.

Part (b) Explanation

The balanced chemical reaction when acetylene burns with air is as follows:

$2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{∆}{\to }4{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

From the above reaction, it is observed that 2 moles of acetylene react with five moles of oxygen, to produce four moles of carbon dioxide and two moles of water.

Therefore, to calculate the number of moles of carbon dioxide, the following conversion factor is required.

$\frac{4\mathrm{moles}{\mathrm{CO}}_2}{2\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}\mathrm{and}\frac{2\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}{4\mathrm{moles}{\mathrm{CO}}_2}$

So, the number of moles of CO₂ that are produced when 3.5 moles of C₂H₂ react is given as follows:

$3.5\cancel{\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}\times \frac{4\mathrm{moles}{\mathrm{CO}}_2}{2\cancel{\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}}=3.5\times 2\mathrm{moles}{\mathrm{CO}}_2=7\mathrm{moles}{\mathrm{CO}}_2$

Hence, 7moles of CO₂ are produced when 3.5 moles of C₂H₂ react .

Part (a) Given Information

The moles ofH₂O are 0.50 moles.

Part (c) Explanation

The balanced chemical reaction when acetylene burns with air is as follows:

$2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{∆}{\to }4{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

From the above reaction, it is observed that 2 moles of acetylene react with five moles of oxygen, to produce four moles of carbon dioxide and two moles of water.

Therefore, for the moles ofC₂H₂ are needed to produce 0.50 moles of H₂O , the following conversion factor is required.

$\frac{2\mathrm{moles}{\mathrm{H}}_2\mathrm{O}}{2\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}\mathrm{and}\frac{2\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}{2\mathrm{moles}{\mathrm{H}}_2\mathrm{O}}$

So, The number of moles ofC₂H₂ that are needed to produce 0.50 moles of H₂O are given as follows:

$0.50\cancel{\mathrm{moles}{\mathrm{H}}_2\mathrm{O}}\times \frac{\cancel{2}\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2}{\cancel{2}\cancel{\mathrm{moles}{\mathrm{H}}_2\mathrm{O}}}=0.50\mathrm{moles}{\mathrm{C}}_2{\mathrm{H}}_2$

Hence. 0.50moles of C₂H₂ are needed to produce 0.50 moles of H₂O .

Part (a) Given Information

The moles ofO₂are 0.200 moles.

Part (d) Explanation

The balanced chemical reaction when acetylene burns with air is as follows:

$2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{∆}{\to }4{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

From the above reaction, it is observed that 2 moles of acetylene react with five moles of oxygen, to produce four moles of carbon dioxide and two moles of water.

Therefore, for the moles of carbon dioxide that is required to produce from 0.200 moles of O₂ , the following conversion factor takes place:

$\frac{5\mathrm{moles}{\mathrm{O}}_2}{4\mathrm{moles}{\mathrm{CO}}_2}\mathrm{and}\frac{4\mathrm{moles}{\mathrm{CO}}_2}{5\mathrm{moles}{\mathrm{O}}_2}$

So, the moles of CO₂ that are produced from 0.200 moles of O₂ are given as follows:

$0.200\mathrm{m}\cancel{\mathrm{oles}{\mathrm{O}}_2}\times \frac{4\mathrm{moles}{\mathrm{CO}}_2}{5\cancel{\mathrm{moles}{\mathrm{O}}_2}}=0.200\times 0.8=0.16\mathrm{moles}{\mathrm{CO}}_2$

Hence, 0.16moles of CO₂ are produced from 0.200 moles of O₂.