Scan_Doc0019 - to react with 1.06 mol NH 3 Since only 1.14...

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112 Chapter Three Stoichiometry What do we know? ./ The chemical reaction NH3(g) + CuO(s) ~ N 2 (g) + Cu(s) + H 2 0(g) ./ 18.1 g NH3 ./ 90.4 g CuO What information do we need? ./ Balanced equation for the reaction ./ Moles of NH3 ./ Moles of CuO How do we get there? To find the limiting reactant What is the balanced equation? 2NH3(g) + 3CuO(s) ~ N 2 (g) + 3Cu(s) + 3H 2 0(g) What are the moles of NH3 and CuO? To find the moles, we need to know the molar masses. NH3 17.03 g/mol CuO 79.55 g/mol 1 mol NH3 = 1.06 mol NH3 18.1 g-NH'3 X 17.03 g-NH'3 1 mol CuO = 1.14 mol CuO 90.4 g. .QnO X 79.55 g. .QnO What is the mole ratio between NH3 and CuO in the balanced equation? 3 mol CuO 2 mol NH3 How many moles of CuO are required to react with 1.06 mol NH3? 3 mol CuO 1.06 mGl-NH 3 X = 1.59 mol CuO 2 JllGl.-NH3 Thus 1.59 mol CuO is required
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Unformatted text preview: to react with 1.06 mol NH 3 . Since only 1.14 mol CuO is actually present, the amount of CuO is limiting; CuO will run out before NH3 does. We can verify this conclusion by comparing the mole ratio of CuO and NH3 required by the balanced equation: molCuO 3---(required) = - = 1.5 molNH3 2 with the mole ratio actually present: mol CuO 1.14---(actual) = - = 1.08 molNH3 1.06 • Since the actual ratio is too small (smaller than 1.5), CuO is the limiting reactant. To find the mass of N2 produced What are the moles of N2 formed? Because CuO is the limiting reactant, we must use the amount of CuO to calculate the amount of N2 formed....
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