Question

For the following reaction, 19.0 grams of nitrogen monoxide are allowed to react with 5.17 grams of hydrogen gas. nitrogen monoxide(g) + hydrogen(g) —nitrogen(g) + water(1) What is the maximum amount of nitrogen gas that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

The Balanced equation is 2 NO(g) + 2 H2(g) ----> N2(g) + 2 H20 (E) Mass of Molar Mass of Molar Mass of Moles of NO NO N2 NO = = 19.00 g 30.010 g/mol 28.00 g/mol Mass = Molar Mass 19.000 g 30.01 = g/mol = 0.633 mol Mass of Molar Mass of Moles of H2 H2 H2 = = 5.17 g 2.016 g/mol Mass - Molar Mass 5.170 g 2.02 g/mol = 2.564 mol i. To Identify the Limiting Reactant: Moles of N2 From = NO is mol of NO 1 mole 2 mole 1 mole 2 mole N2 NO N2 NO 0.6331 . moles NO = 0.3166 moles N2 Moles of N2 From * H2 * 1 N2 mol of mole H2 is 2 mole H2 15 2.5645 moles i mole N2 H2 2 mole H2 1.282 moles N2 NO Produce less moles of N2 SO NO is limiting Reactant H2 is Excess Reagent ii. Calculation of Theoretical yield: Theoretical yield of N2 Moles * Molar mass = 0.3166 mol* 28.00 g/mol = 8.86 8 iv. Calculation of mass of Excess reagent: Moles of H2 moles 2 mol H2 Consumed in Reaction NO 2 mol NO = 0.633 mol H2 Moles of H2 excess Moles H2 consumed Actual moles of H2 2.564 - 0.6331 1.931 mol H2 Moles * Molar mass 1.9314 mol H2 3.898 Mass of H2 excess * 2.016 g/mol