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 resultant 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 balanced chemical equation for the combustion of acetylene (C2H2) in the presence of oxygen (O2) forming carbon dioxide (CO2) and water (H2O) is: \[2 C_{2}H_{2} + 5 O_{2} \rightarrow 4 CO_{2} + 2 H_{2}O\]

To determine the limiting reactant, we will first calculate the moles of each reactant present: Given that there are 10 grams of each C2H2 and O2, Moles of C2H2 = \(\frac{10.0 \ \text{g}}{26.04 \ \text{g/mol}} = 0.384 \ \text{moles}\) Moles of O2 = \(\frac{10.0 \ \text{g}}{32.00 \ \text{g/mol}} = 0.313 \ \text{moles}\)

Now, we will check which reactant is the limiting reactant: Dividing the moles of each reactant by its respective stoichiometric coefficient: C2H2: \(\frac{0.384 \ \text{moles}}{2} = 0.192\) O2: \(\frac{0.313 \ \text{moles}}{5} = 0.063\) Since the value for O2 (0.063) is smaller than the value for C2H2 (0.192), oxygen is the limiting reactant.

Using the limiting reactant, we can calculate the mass of each substance present after the reaction: (a) Mass of remaining C2H2: Since 5 moles of O2 react with 2 moles of C2H2, \(\ 0.313\ \text{moles} \times \frac{2 \ \text{moles} \ C2H2}{5 \ \text{moles} \ O2} = 0.126\ \text{moles} \ \text{of consumed C2H2}\) Remaining C2H2 = Initial C2H2 - Consumed C2H2 = \(0.384 \ \text{moles} - 0.126 \ \text{moles} = 0.258 \ \text{moles}\) Mass of remaining C2H2 = \(0.258 \ \text{moles} \times 26.04 \ \text{g/mol} = 6.71\ \text{g}\) (b) Mass of remaining O2: Since O2 is the limiting reactant, it must be fully consumed, so the mass of remaining O2 is \(0\ \text{g}\). (c) Mass of generated CO2: Given 2 moles of C2H2 react with 5 moles of O2 to form 4 moles of CO2, \(\ 0.313\ \text{moles} \times \frac{4 \ \text{moles} \ CO2}{5 \ \text{moles} \ O2} = 0.251\ \text{moles} \ \text{of CO2}\) Mass of generated CO2 = \(0.251\ \text{moles} \times 44.01\ \text{g/mol} = 11.04\ \text{g}\) (d) Mass of generated H2O: Given 2 moles of C2H2 react with 5 moles of O2 to form 2 moles of H2O, \(\ 0.313\ \text{moles} \times \frac{2 \ \text{moles} \ H2O}{5 \ \text{moles} \ O2} = 0.126\ \text{moles} \ \text{of H2O}\) Mass of generated H2O = \(0.126 \ \text{moles} \times 18.02\ \text{g/mol} = 2.27\ \text{g}\)

After the combustion reaction is complete: - Mass of remaining C2H2: 6.71 g - Mass of remaining O2: 0 g - Mass of generated CO2: 11.04 g - Mass of generated H2O: 2.27 g