The combustion of liquid chloroethylene, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), yields carbon dioxide, steam, and hydrogen chloride gas. (a) Write a balanced equation for the reaction. (b) How many moles of oxygen are required to react with \(35.00 \mathrm{~g}\) of chloroethylene? (c) If \(25.00 \mathrm{~g}\) of chloroethylene reacts with an excess of oxygen, how many grams of each product are formed?

The given chloroethylene is \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\). In a combustion reaction, the reactants are the substance being combusted (in this case, chloroethylene) and oxygen gas (O₂). The products are carbon dioxide (CO₂), steam (H₂O), and hydrogen chloride gas (HCl). The unbalanced equation is: $$\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O} + \mathrm{HCl}$$ Now, we need to balance the equation. We can do this by changing the coefficients in front of each compound until the number of atoms of each element is the same on both sides of the equation: $$\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl} + \frac{7}{2} \mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O} + 2\mathrm{HCl}$$

First, we need to find the molar mass of chloroethylene (C₂H₃Cl): Molar mass of C₂H₃Cl = (2 * molar mass of C) + (3 * molar mass of H) + (1 * molar mass of Cl) Molar mass of C₂H₃Cl = (2 * 12.01 g/mol) + (3 * 1.01 g/mol) + (1 * 35.45 g/mol) = 62.97 g/mol Now, we can find the moles of chloroethylene given 35.00 g: Moles of C₂H₃Cl = mass / molar mass = 35.00 g / 62.97 g/mol = 0.555 mol Using the balanced equation, we can find the moles of oxygen required: Moles of O₂ = moles of C₂H₃Cl * (7/2 moles O₂ / 1 mole C₂H₃Cl) = 0.555 mol * (7/2) = 1.943 mol

First, we need to find the moles of chloroethylene given 25.00 g: Moles of C₂H₃Cl = mass / molar mass = 25.00 g / 62.97 g/mol = 0.397 mol Using the balanced equation, we can find the moles of each product formed: Moles of CO₂ = moles of C₂H₃Cl * (2 moles CO₂ / 1 mole C₂H₃Cl) = 0.397 mol * 2 = 0.794 mol Moles of H₂O = moles of C₂H₃Cl * (1 mole H₂O / 1 mole C₂H₃Cl) = 0.397 mol * 1 = 0.397 mol Moles of HCl = moles of C₂H₃Cl * (2 moles HCl / 1 mole C₂H₃Cl) = 0.397 mol * 2 = 0.794 mol Now we can find the mass of each product formed using their respective molar masses: Mass of CO₂ = moles of CO₂ * molar mass of CO₂ = 0.794 mol * 44.01 g/mol = 34.92 g Mass of H₂O = moles of H₂O * molar mass of H₂O = 0.397 mol * 18.02 g/mol = 7.15 g Mass of HCl = moles of HCl * molar mass of HCl = 0.794 mol * 36.46 g/mol = 28.95 g The grams of each product formed are 34.92 g of CO₂, 7.15 g of H₂O, and 28.95 g of HCl.