d. The mole of HNO3 needed and the total mole of HNO3 available has already been found in part a of the problem. To find the mole of HNO3 in excess, subtract the mole of HNO3 needed from the total mole of HNO3 available: 1.12 mol HNO3 0.924 mol HNO3 = 0.20 mol HNO3 in excess – e. A percent is always a ratio. Use the actual yield as the and the theoretical “part/whole” “part” yield as the “whole”:
d. The mole of HNO3 needed and the total mole of HNO3 available has already been found in part a of the problem. To find the mole of HNO3 in excess, subtract the mole of HNO3 needed from the total mole of HNO3 available: 1.12 mol HNO3 – 0.924 mol HNO3 = 0.20 mol HNO3 in excess e. A percent is always a “part/whole” ratio. Use the actual yield as the “part” and the theoretical yield as the “whole”:
akstoich.docx