Final answer:
The balanced equation for the combustion of acetylene is 2C2H2 + 5O2 → 4CO2 + 2H2O. Oxygen is the limiting reactant, producing 13.73 g of CO2 and 2.81 g of H2O, with 8.42 g of acetylene remaining unreacted.
Step-by-step explanation:
Let's tackle a combustion analysis of acetylene (C2H2) and oxygen (O2). We start by writing the balanced chemical equation for the combustion reaction:
2C2H2 (g) + 5O2 (g) → 4CO2 (g) + 2H2O (l)
To determine the limiting reactant, we must calculate the moles of each reactant. Acetylene (C2H2) has a molar mass of approximately 26 g/mol, so 12.5 g corresponds to about 0.48 moles. Oxygen (O2) has a molar mass of 32 g/mol, hence 12.5 g is roughly 0.39 moles. According to the balanced equation, the stoichiometry requires 5 moles of O2 for every 2 moles of C2H2. Thus, oxygen is the limiting reactant because we have less than the required proportion compared to acetylene.
Since oxygen is the limiting reactant, we calculate molar amounts of CO2 and H2O produced based on the moles of oxygen. With 0.39 moles of O2, we would produce 0.39*4/5 = 0.312 moles of CO2 and 0.39*2/5 = 0.156 moles of H2O. The mass of CO2 produced is 0.312 moles * 44 g/mol = 13.73 g, and the mass of H2O produced is 0.156 moles * 18 g/mol = 2.81 g.
To find the excess reactant remaining, we calculate the unreacted acetylene. Initially, we had 0.48 moles of C2H2, and the reaction used 0.39*2/5 = 0.156 moles, leaving 0.48 - 0.156 = 0.324 moles of C2H2 unreacted. This corresponds to 0.324 moles * 26 g/mol = 8.42 g of acetylene remaining.