A certain sample of coal contains 1.60 percent sulfur by mass. when the coal is burned, the sulfur is converted to sulfur dioxide. to prevent air pollution, this sulfur dioxide is treated with calcium oxide (cao) to form calcium sulfite (caso3). calculate the daily mass (in kilograms) of cao needed by a power plant that uses 5.00 × 106 kg of coal per day. enter your answer in scientific notation.

1.40 x 10^5 kilograms of calcium oxide The reaction looks like SO2 + CaO => CaSO3 First, determine the mass of sulfur in the coal 5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4 Now lookup the atomic weights of Sulfur, Calcium, and Oxygen. Sulfur = 32.065 Calcium = 40.078 Oxygen = 15.999 Calculate the molar mass of CaO CaO = 40.078 + 15.999 = 56.077 Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight. 8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass 2.49 x 10^3 * 56.077 = 1.40 x 10^5 So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.