When a mixture of \(10.0 \mathrm{~g}\) of acetylene $\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\( and \)10.0 \mathrm{~g}$ of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resulting combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of $\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2},\( and \)\mathrm{H}_{2} \mathrm{O}$ are present after the reaction is complete?

The combustion reaction of acetylene \(\left(\mathrm{C}_{2}\mathrm{H}_{2}\right)\) with oxygen \(\left(\mathrm{O}_{2}\right)\) produces carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and water \(\left(\mathrm{H}_{2}\mathrm{O}\right)\). First, write an unbalanced chemical equation. \[\mathrm{C}_{2}\mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\] Next, balance the chemical equation. There are two carbons, two hydrogens, and two oxygens on the left side and one carbon, two hydrogens, and three oxygens on the right side. The balanced chemical equation is: \[\mathrm{C}_{2}\mathrm{H}_{2} + 2.5\mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\] However, fractions in stoichiometry are unusual, so it's preferable to use whole numbers. To achieve this, multiply the entire reaction by 2: \[2\mathrm{C}_{2}\mathrm{H}_{2} + 5\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\] This balanced chemical equation represents the reaction.

To determine the limiting reactant, we need to know the amount of each reactant in moles. Use the molar mass to convert grams to moles. The molar mass of \(\mathrm{C}_{2}\mathrm{H}_{2}\) is 26.04 g/mol, and that of \(\mathrm{O}_{2}\) is 32.00 g/mol. For \(\mathrm{C}_{2}\mathrm{H}_{2}\): \[10.0 \mathrm{~g}\, \mathrm{C}_{2}\mathrm{H}_{2} \times \frac{1\, \text{mol} \, \mathrm{C}_{2}\mathrm{H}_{2}}{26.04\, \mathrm{g}\, \mathrm{C}_{2}\mathrm{H}_{2}} \approx 0.384\, \mathrm{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}\] For \(\mathrm{O}_{2}\): \[10.0 \mathrm{~g}\, \mathrm{O}_{2} \times \frac{1\, \text{mol}\, \mathrm{O}_{2}}{32.00\, \mathrm{g}\, \mathrm{O}_{2}} \approx 0.312\, \text{mol}\, \mathrm{O}_{2}\] Now, calculate the mole ratios by dividing the moles of each substance by its stoichiometric coefficient: For \(\mathrm{C}_{2}\mathrm{H}_{2}\): \(\frac{0.384}{2} = 0.192\) For \(\mathrm{O}_{2}\): \(\frac{0.312}{5} = 0.062\) The smallest mole ratio value corresponds to the limiting reactant, which is \(\mathrm{O}_{2}\).

Since oxygen is the limiting reactant, we can calculate the amount of acetylene remaining and the amounts of carbon dioxide and water formed using the balanced equation. 1. Calculate moles of acetylene remaining: From the balanced equation, 2 moles of \(\mathrm{C}_{2}\mathrm{H}_{2}\) react with 5 moles of \(\mathrm{O}_{2}\). Therefore, \(0.312\, \text{mol}\, \mathrm{O}_{2}\) will react with: \[0.312\, \text{mol}\, \mathrm{O}_{2} \times \frac{2\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}}{5\, \text{mol}\, \mathrm{O}_{2}} \approx 0.125\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}\] Subtract the reacted \(\mathrm{C}_{2}\mathrm{H}_{2}\) from the initial amount: \[0.384\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2} - 0.125\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2} \approx 0.259\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}\] Convert moles to grams: \[0.259\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2} \times 26.04\, \mathrm{g}\, \mathrm{C}_{2}\mathrm{H}_{2}\, \text{mol}^{-1} \approx 6.74\, \mathrm{g}\, \mathrm{C}_{2}\mathrm{H}_{2}\] 2. Calculate moles and grams of carbon dioxide produced: From the balanced equation, 4 moles of \(\mathrm{CO}_{2}\) are produced for every 5 moles of \(\mathrm{O}_{2}\) reacted: \[0.312\, \text{mol}\, \mathrm{O}_{2} \times \frac{4\, \text{mol}\, \mathrm{CO}_{2}}{5\, \text{mol}\, \mathrm{O}_{2}} \approx 0.250\, \text{mol}\, \mathrm{CO}_{2}\] Convert moles to grams: \[0.250\, \text{mol}\, \mathrm{CO}_{2} \times 44.01\, \mathrm{g}\, \mathrm{CO}_{2}\, \text{mol}^{-1} \approx 11.0\, \mathrm{g}\, \mathrm{CO}_{2}\] 3. Calculate moles and grams of water produced: From the balanced equation, 2 moles of \(\mathrm{H}_{2}\mathrm{O}\) are produced for every 5 moles of \(\mathrm{O}_{2}\) reacted: \[0.312\, \mathrm{mol}\, \mathrm{O}_{2} \times \frac{2\, \text{mol}\, \mathrm{H}_{2}\mathrm{O}}{5\, \text{mol}\, \mathrm{O}_{2}} \approx 0.125\, \text{mol}\, \mathrm{H}_{2}\mathrm{O}\] Convert moles to grams: \[0.125\, \text{mol}\, \mathrm{H}_{2}\mathrm{O} \times 18.02\, \mathrm{g}\, \mathrm{H}_{2}\mathrm{O}\, \text{mol}^{-1} \approx 2.25\, \mathrm{g}\, \mathrm{H}_{2}\mathrm{O}\] Final amounts after the reaction is complete: \(\mathrm{C}_{2}\mathrm{H}_{2}\): \(6.74\, \mathrm{g}\) \(\mathrm{O}_{2}\): \(0\, \mathrm{g}\) (limiting reactant, all reacted) \(\mathrm{CO}_{2}\): \(11.0\, \mathrm{g}\) \(\mathrm{H}_{2}\mathrm{O}\): \(2.25\, \mathrm{g}\)