18.1g of ammonia and 90.4g of copper(II) oxide are used in the reaction 2NH3+3CuO=N2+3Cu+3H3O.What is the limiting reactant? How many grams of nitrogen will be formed?
To determine the limiting reactant and calculate the amount of nitrogen formed, we need to compare the amount of product that can be formed from each reactant.
1. Calculate the number of moles for each reactant:
- Moles of ammonia (NH3) = mass of ammonia / molar mass of ammonia
- Moles of copper(II) oxide (CuO) = mass of copper(II) oxide / molar mass of copper(II) oxide
2. Determine the stoichiometric ratio between reactants and products:
- The balanced equation is 2NH3 + 3CuO = N2 + 3Cu + 3H2O.
- From the equation, we can see that 2 moles of ammonia react with 3 moles of copper(II) oxide to produce 1 mole of nitrogen.
3. Calculate the maximum moles of nitrogen that can be produced from each reactant:
- Moles of nitrogen from ammonia = moles of ammonia * (1 mole of nitrogen / 2 moles of ammonia)
- Moles of nitrogen from copper(II) oxide = moles of copper(II) oxide * (1 mole of nitrogen / 3 moles of copper(II) oxide)
4. Compare the moles of nitrogen produced from each reactant:
- The reactant that produces the lesser amount of moles of nitrogen is the limiting reactant.
5. Convert the moles of limiting reactant to grams of nitrogen:
- Grams of nitrogen = moles of limiting reactant * molar mass of nitrogen
Let's do the calculations:
1. Moles of ammonia:
Molar mass of ammonia (NH3) = 14.01 g/mol + 1.01 g/mol (3 hydrogen atoms) = 17.03 g/mol
Moles of ammonia = 18.1 g / 17.03 g/mol ≈ 1.064 mol
2. Moles of copper(II) oxide:
Molar mass of copper(II) oxide (CuO) = 63.55 g/mol + 16.00 g/mol = 79.55 g/mol
Moles of copper(II) oxide = 90.4 g / 79.55 g/mol ≈ 1.136 mol
3. Moles of nitrogen from ammonia:
Moles of nitrogen from ammonia = 1.064 mol * (1 mol N2 / 2 mol NH3) ≈ 0.532 mol
Moles of nitrogen from copper(II) oxide:
Moles of nitrogen from copper(II) oxide = 1.136 mol * (1 mol N2 / 3 mol CuO) ≈ 0.379 mol
4. Comparing moles of nitrogen formed:
The moles of nitrogen formed from ammonia (0.532 mol) is greater than the moles of nitrogen formed from copper(II) oxide (0.379 mol). Therefore, ammonia is the limiting reactant.
5. Grams of nitrogen formed:
Grams of nitrogen = 0.532 mol * 28.01 g/mol ≈ 14.9 g
Therefore, the limiting reactant is ammonia, and approximately 14.9 grams of nitrogen will be formed.
To determine the limiting reactant, we need to compare the amount of product each reactant can produce.
1. Calculate the number of moles of ammonia and copper(II) oxide:
moles of ammonia = mass / molar mass = 18.1g / 17.03g/mol ≈ 1.0647mol
moles of copper(II) oxide = mass / molar mass = 90.4g / 79.55g/mol ≈ 1.1363mol
2. Determine the stoichiometric ratio between reactants and product:
From the balanced equation, we can see that 2 moles of NH3 react with 3 moles of CuO to produce 1 mole of N2.
3. Calculate the moles of nitrogen produced from each reactant:
moles of nitrogen from ammonia = (1.0647 mol NH3) / (2 mol NH3/1 mol N2) ≈ 0.53235 mol N2
moles of nitrogen from copper(II) oxide = (1.1363 mol CuO) / (3 mol CuO/1 mol N2) ≈ 0.37877 mol N2
4. Determine the limiting reactant:
The reactant that produces fewer moles of nitrogen is the limiting reactant. In this case, copper(II) oxide produces 0.37877 mol N2, which is less than the 0.53235 mol N2 produced by ammonia.
Therefore, the limiting reactant is copper(II) oxide.
5. Calculate the grams of nitrogen formed:
mass of nitrogen = moles of nitrogen × molar mass = 0.37877 mol × 28.013g/mol ≈ 10.613g
Approximately, 10.613 grams of nitrogen will be formed.
mols NH3 = grams/molar mass = estimated 1.0
mols CuO = g/molar mass = estimated 1.1
Convert mols each to mols product. N2 is the easiest.
1 mol NH3 x (1 mol N2/2 mol NH3) = 1 x 1/2 = estimated 1/2 mol N2.
1.1 mol CuO x (1 mol N2/3 mol CuO) = estimated 1.1 x 1/3 = about 0.4
So CuO is the limiting reagent because the smaller amount will be formed.
g N2 = mols N2 x molar mass N2.