If 120 g of propane, C_3H_8, is burned in excess oxygen according to the following equation: C_3H_8 + 5O_2 to 3CO_2 + 4H_2O, how many grams of water are formed?

According to the balanced chemical equation, 1 mole of propane (C3H8) reacts with 4 moles of oxygen (O2) to produce 4 moles of water (H2O). First, we need to calculate the number of moles of propane in 120 g of C3H8: molar mass of C3H8 = (3 x 12.01 g/mol) + (8 x 1.01 g/mol) = 44.1 g/mol moles of C3H8 = mass / molar mass = 120 g / 44.1 g/mol = 2.72 moles Next, we need to determine the number of moles of water produced from 2.72 moles of propane: moles of water = 4 x moles of propane = 4 x 2.72 moles = 10.88 moles Finally, we can calculate the mass of water produced from the number of moles: mass of water = moles of water x molar mass of water molar mass of water = 2 x 1.01 g/mol + 1 x 16.00 g/mol = 18.02 g/mol mass of water = 10.88 moles x 18.02 g/mol = 196 g Therefore, 196 grams of water are formed when 120 g of propane is burned in excess oxygen.