Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperature. The other products of the reaction are solid copper and water vapor. If a sample containing 20.3 g of NH3 is reacted with 95.8 g of copper oxide
1. Which is the liming reactant?
2. How many moles of nitrogen gas is being produced?
3. How many grams of solid copper?
4. How many grams of water?
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To determine the limiting reactant, we need to compare the moles of each reactant to the stoichiometry of the balanced equation. The balanced equation for the reaction is:
2 NH3 + 3 CuO -> N2 + 3 Cu + 3 H2O
1. To find the limiting reactant, convert the given mass of each reactant to moles.
Molar mass of NH3 = 17.03 g/mol
Molar mass of CuO = 79.55 g/mol
Number of moles of NH3 = 20.3 g / 17.03 g/mol = 1.19 mol
Number of moles of CuO = 95.8 g / 79.55 g/mol = 1.20 mol
Based on the stoichiometry of the balanced equation, it can be observed that 2 moles of NH3 react with 3 moles of CuO. Thus, NH3 is the limiting reactant, as there are fewer moles of NH3 compared to CuO.
2. The balanced equation indicates that 2 moles of NH3 react to produce 1 mole of N2. Therefore, the number of moles of nitrogen gas produced can be determined by the number of moles of NH3 reacted.
Number of moles of N2 produced = 1.19 mol NH3 * (1 mol N2 / 2 mol NH3) = 0.595 mol
3. According to the balanced equation, if 2 moles of NH3 react, 3 moles of Cu are produced. Thus, we can determine the number of moles of Cu produced using the stoichiometry of the reaction.
Number of moles of Cu produced = 0.595 mol N2 * (3 mol Cu / 1 mol N2) = 1.79 mol
To find the mass of solid copper produced, we can use its molar mass:
Molar mass of Cu = 63.55 g/mol
Mass of solid copper produced = 1.79 mol * 63.55 g/mol = 113.9 g
4. Finally, we can determine the mass of water produced using the stoichiometry of the reaction.
According to the balanced equation, if 2 moles of NH3 react, 3 moles of H2O are produced.
Number of moles of H2O produced = 0.595 mol N2 * (3 mol H2O / 2 mol NH3) = 0.893 mol
To find the mass of water produced, we can use its molar mass:
Molar mass of H2O = 18.02 g/mol
Mass of water produced = 0.893 mol * 18.02 g/mol = 16.09 g
In conclusion:
1. The limiting reactant is NH3.
2. The number of moles of nitrogen gas produced is 0.595 mol.
3. The mass of solid copper produced is 113.9 g.
4. The mass of water produced is 16.09 g.