Questions: Acetylene gas C2H2 undergoes combustion to form carbon dioxide and water when it is used in the oxyacetylene torch for welding. Balance the reaction and answer the following questions. C2H2(g) + O2(g) -> CO2(g) + H2O(g) (a) How many grams of water can form if 113 g of acetylene is burned? (b) How many grams of acetylene react if 1.10 mol of CO2 are produced?

Acetylene gas C2H2 undergoes combustion to form carbon dioxide and water when it is used in the oxyacetylene torch for welding. Balance the reaction and answer the following questions.
C2H2(g) + O2(g) -> CO2(g) + H2O(g)
(a) How many grams of water can form if 113 g of acetylene is burned?
(b) How many grams of acetylene react if 1.10 mol of CO2 are produced?

Transcript text: 2. Acetylene gas $\mathrm{C}_{2} \mathrm{H}_{2}$ undergoes combustion to form carbon dioxide and water when it is used in the oxyacetylene torch for welding. Balance the reaction and answer the following questions. \[ \mathrm{C}_{2} \mathrm{H}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \] (a) How many grams of water can form if 113 g of acetylene is burned? (b) How many grams of acetylene react if 1.10 mol of $\mathrm{CO}_{2}$ are produced?

Solution

Solution Steps

Step 1: Balance the Chemical Equation

First, balance the chemical equation for the combustion of acetylene: \[ \mathrm{C}_{2} \mathrm{H}_{2(\mathrm{~g})} + \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})} + \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \]

Balanced equation: \[ 2 \mathrm{C}_{2} \mathrm{H}_{2(\mathrm{~g})} + 5 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 4 \mathrm{CO}_{2(\mathrm{~g})} + 2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \]

Step 2: Calculate Molar Masses

Calculate the molar masses of the relevant compounds:

Molar mass of $\mathrm{C}_{2} \mathrm{H}_{2}$: $2 \times 12.01 + 2 \times 1.01 = 26.04 \, \text{g/mol}$
Molar mass of $\mathrm{H}_{2} \mathrm{O}$: $2 \times 1.01 + 16.00 = 18.02 \, \text{g/mol}$

Step 3: Convert Grams of Acetylene to Moles

Convert 113 g of acetylene to moles: \[ \text{Moles of } \mathrm{C}_{2} \mathrm{H}_{2} = \frac{113 \, \text{g}}{26.04 \, \text{g/mol}} \approx 4.34 \, \text{mol} \]

Step 4: Use Stoichiometry to Find Moles of Water

Use the balanced equation to find the moles of water produced: \[ 2 \, \text{mol} \, \mathrm{C}_{2} \mathrm{H}_{2} \rightarrow 2 \, \text{mol} \, \mathrm{H}_{2} \mathrm{O} \] \[ 4.34 \, \text{mol} \, \mathrm{C}_{2} \mathrm{H}_{2} \rightarrow 4.34 \, \text{mol} \, \mathrm{H}_{2} \mathrm{O} \]

Step 5: Convert Moles of Water to Grams

Convert moles of water to grams: \[ \text{Grams of } \mathrm{H}_{2} \mathrm{O} = 4.34 \, \text{mol} \times 18.02 \, \text{g/mol} \approx 78.20 \, \text{g} \]

Step 6: Convert Moles of $\mathrm{CO}_{2}$ to Moles of Acetylene

Convert 1.10 mol of $\mathrm{CO}_{2}$ to moles of acetylene using the balanced equation: \[ 4 \, \text{mol} \, \mathrm{CO}_{2} \rightarrow 2 \, \text{mol} \, \mathrm{C}_{2} \mathrm{H}_{2} \] \[ 1.10 \, \text{mol} \, \mathrm{CO}_{2} \rightarrow \frac{1.10 \, \text{mol} \times 2}{4} = 0.55 \, \text{mol} \, \mathrm{C}_{2} \mathrm{H}_{2} \]

Step 7: Convert Moles of Acetylene to Grams

Convert moles of acetylene to grams: \[ \text{Grams of } \mathrm{C}_{2} \mathrm{H}_{2} = 0.55 \, \text{mol} \times 26.04 \, \text{g/mol} \approx 14.32 \, \text{g} \]