Sample Problem 3.10 Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem. continued.
{"Image DocumentName": "Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem. continued.", "Image ImageDescription": "SOLUTION: 2. N2H4(l) + N2O4(l) N2(g) + H2O(l) mol N2H g N2H x102g N2H4. = 3.12mol N2H4. N2H4 is the limiting reactant because it produces less product, N2, than does N2O4. 3 mol N2. 2mol N2H mol N2H4. = 4.68mol N2. mol N g N mol N2. = 131g N2. mol N2O g N2O x102g N2O4. = 2.17mol N2O4. 3 mol N2. mol N2O mol N2O4. = 6.51mol N2.", "Image XResolution": "37", "Image YResolution": "37", "Image ResolutionUnit": "Pixels/Centimeter", "Image Software": "SlidePlayer", "Image YCbCrPositioning": "Centered"}