Question

In the acetylene torch, acetylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) burns in oxygen to produce carbon dioxide and water. $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ a. How many moles of \(\mathrm{O}_{2}\) are needed to react with \(2.00\) moles of \(\mathrm{C}_{2} \mathrm{H}_{2} ?\) b. How many moles of \(\mathrm{CO}_{2}\) are produced when \(3.5\) moles of \(\mathrm{C}_{2} \mathrm{H}_{2}\) reacts? c. How many moles of \(\mathrm{C}_{2} \mathrm{H}_{2}\) are required to produce \(0.50\) mole of \(\mathrm{H}_{2} \mathrm{O} ?\) d. How many moles of \(\mathrm{CO}_{2}\) are produced from \(0.100\) mole of \(\mathrm{O}_{2}\) ?

Short answer

a. 5.00 moles of O2. b. 7.0 moles of CO2. c. 0.50 moles of C2H2. d. 0.080 moles of CO2.

Step-by-step solution

Title - Balance the chemical equation

The given chemical equation is already balanced: \[2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\]

Title - Calculate moles of O2 needed (Part a)

From the balanced equation, 2 moles of \mathrm{C}_{2} \mathrm{H}_{2} react with 5 moles of \mathrm{O}_{2}. Therefore, if you have 2.00 moles of \mathrm{C}_{2} \mathrm{H}_{2}, then the moles of \mathrm{O}_{2} needed: \[2.00 \text{ moles } \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{5 \text{ moles } \mathrm{O}_{2}}{2 \text{ moles } \mathrm{C}_{2} \mathrm{H}_{2}}=5.00 \text{ moles } \mathrm{O}_{2}\]

Title - Calculate moles of CO2 produced (Part b)

From the balanced equation, 2 moles of \mathrm{C}_{2} \mathrm{H}_{2} produce 4 moles of \mathrm{CO}_{2}. Given 3.5 moles of \mathrm{C}_{2} \mathrm{H}_{2}: \[3.5 \text{ moles } \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{4 \text{ moles } \mathrm{CO}_{2}}{2 \text{ moles } \mathrm{C}_{2} \mathrm{H}_{2}}=7.0 \text{ moles } \mathrm{CO}_{2}\]

Title - Calculate moles of C2H2 required (Part c)

From the balanced equation, 2 moles of \mathrm{C}_{2} \mathrm{H}_{2} produce 2 moles of \mathrm{H}_{2} \mathrm{O}. Therefore, the moles of \mathrm{C}_{2} \mathrm{H}_{2} needed to produce 0.50 moles of \mathrm{H}_{2} \mathrm{O} are: \[0.50 \text{ moles } \mathrm{H}_{2} \mathrm{O} \times \frac{2 \text{ moles } \mathrm{C}_{2} \mathrm{H}_{2}}{2 \text{ moles } \mathrm{H}_{2} \mathrm{O}}=0.50 \text{ moles } \mathrm{C}_{2} \mathrm{H}_{2}\]

Title - Calculate moles of CO2 produced from O2 (Part d)

From the balanced equation, 5 moles of \mathrm{O}_{2} produce 4 moles of \mathrm{CO}_{2}. Therefore, the moles of \mathrm{CO}_{2} produced from 0.100 mole of \mathrm{O}_{2} are: \[0.100 \text{ moles } \mathrm{O}_{2} \times \frac{4 \text{ moles } \mathrm{CO}_{2}}{5 \text{ moles } \mathrm{O}_{2}}=0.080 \text{ moles } \mathrm{CO}_{2}\]

Key concepts

chemical reactions

Chemical reactions are processes where reactants transform into products. In our exercise, acetylene gas \(\text{C}_2\text{H}_2\) reacts with oxygen \(\text{O}_2\), forming carbon dioxide \(\text{CO}_2\) and water \(\text{H}_2\text{O}\). These transformations happen according to the law of conservation of mass, which states that matter cannot be created or destroyed, only rearranged.

mole calculations

Moles are a basic unit in chemistry used to describe the amount of a substance. One mole corresponds to \(6.022 \times 10^{23}\) particles (Avogadro's number). The given exercise involves multiple instances of mole calculations:

• For part a: Moles of \(\text{O}_2\) needed for a given moles of \(\text{C}_2\text{H}_2\) is calculated.
• For part b: Moles of \(\text{CO}_2\) produced from given \(\text{C}_2\text{H}_2\).
• For part c: Moles of \(\text{C}_2\text{H}_2\) needed to produce given moles of \(\text{H}_2\text{O}\).
• For part d: Moles of \(\text{CO}_2\) produced from given moles of \(\text{O}_2\).

These calculations rely on the stoichiometric coefficients from the balanced chemical equation.

balanced equations

A balanced equation has equal numbers of each type of atom on both sides. In our exercise, the balanced equation is:

\[2 \text{C}_2\text{H}_2\text{(g)} + 5 \text{O}_2\text{(g)} \rightarrow 4 \text{CO}_2\text{(g)} + 2 \text{H}_2\text{O}\text{(g)}\]

This ensures the law of conservation of mass is obeyed. In the above reaction:

• 2 molecules of \(C_2H_2\) react with 5 molecules of \(O_2\).
• Producing 4 molecules of \(CO_2\) and 2 molecules of \(H_2O\).

A balanced equation is crucial for calculating the quantities involved in a reaction.

chemical equation balancing

Balancing chemical equations ensures the same number of atoms for each element on both sides of the equation. Follow these steps:

• Write the unbalanced equation.
• List the number of atoms of each element on both sides.
• Adjust coefficients to get the same number of each atom on both sides.
• Ensure the coefficients are in the simplest whole number ratio.

For example, in the given reaction:

• Start with \(C_2H_2 + O_2 \rightarrow CO_2 + H_2O\).
• Balance \(C\) atoms: \(2C_2H_2 \rightarrow 4CO_2\).
• Balance \(H\) atoms: \(2C_2H_2 \rightarrow 4 CO_2 + 2H_2O\).
• Finally, balance \(O\) atoms: \(5 O_2 + 2C_2H_2 \rightarrow 4 CO_2 + 2 H_2O\).