Ammonia reacts with copper(ll)oxide to give nitrogen gas, copper and steam. Balance the equation. If 18.1g of ammonia is reacted with 90.4g of coope(ll)oxide,how many moles of each reacts?,and how many grams of nitrogen are formed?
Cu(II) oxide is CuO
so
3CuO+2NH3 → 3Cu+N2+3H2O
From 2NH3 and 3CuO, we conclude that
2 moles of NH3 reacts with 3 moles of CuO to form 1 mol of N2.
Find the molar masses of NH3 and CuO and determine which is the limiting reagent. Use stoichiometric ratios to calculate the number of moles reacted and the mass of N2 formed.
the limiting reactant is cuo
0.757mole is reacted
7.059g of N2 is produced
To balance the equation, we need to ensure that the number of atoms of each element on both sides of the equation is equal.
The reaction between ammonia and copper(ll) oxide can be represented by the following unbalanced equation:
NH3 + CuO -> N2 + Cu + H2O
Now, let's balance the equation step by step:
1. Start by counting the number of each type of atom on both sides of the equation.
On the left side (reactants):
- Nitrogen: 1
- Hydrogen: 3
- Oxygen: 1
- Copper: 1
On the right side (products):
- Nitrogen: 2
- Hydrogen: 2
- Oxygen: 1
- Copper: 1
2. Begin balancing with the elements that appear in the fewest compounds. In this case, we start with copper.
Adding a coefficient of 2 in front of Cu on the reactant side will balance the number of copper atoms:
NH3 + 2CuO -> N2 + Cu + H2O
3. Now, balance the hydrogen atoms. They are already balanced, so we move on to nitrogen.
Adding a coefficient of 2 in front of NH3 on the reactant side will balance the nitrogen atoms:
2NH3 + 2CuO -> N2 + Cu + H2O
4. Finally, balance the oxygen atoms.
Adding a coefficient of 3 in front of H2O on the product side will balance the oxygen atoms:
2NH3 + 2CuO -> N2 + Cu + 3H2O
The balanced equation is 2NH3 + 2CuO -> N2 + Cu + 3H2O.
Now, let's calculate the number of moles of each reactant and the grams of nitrogen formed.
To calculate the number of moles, we use the formula:
moles = mass / molar mass
1. Ammonia (NH3):
moles = 18.1 g / 17.03 g/mol = 1.0638 mol
2. Copper(ll) oxide (CuO):
moles = 90.4 g / 79.55 g/mol = 1.1366 mol
From the balanced equation, we can see that the stoichiometric ratio between ammonia and nitrogen is 2:1. Therefore, for every 2 moles of ammonia, 1 mole of nitrogen is formed.
Moles of nitrogen = 1.0638 mol / 2 = 0.5319 mol
To calculate the grams of nitrogen formed, we use the formula:
mass = moles * molar mass
Molar mass of nitrogen (N2) = 28.02 g/mol
Mass of nitrogen = 0.5319 mol * 28.02 g/mol = 14.91 g
Therefore, 14.91 grams of nitrogen are formed when 18.1 grams of ammonia react with 90.4 grams of copper(ll) oxide.