Final answer:
By using stoichiometry and the balanced chemical equation for the reaction between ammonia and copper (II) oxide, we calculate that 9.86 grams of nitrogen gas will be formed from 18.0 grams of ammonia.
Step-by-step explanation:
To calculate the amount of nitrogen gas (N2) produced from the reaction of 18.0 grams of ammonia (NH3) with 82.0 grams of copper (II) oxide, we must first write the balanced chemical equation for the reaction. The balanced equation is:
3 NH3 + 3 CuO → N2 + 3 Cu + 3 H2O
Next, we determine the molar mass of NH3 and the molar mass of N2. The molar mass of NH3 is 14.01 (for N) + 3(1.008) (for H) = 17.03 g/mol. The molar mass of N2 is 2(14.01) = 28.02 g/mol.
Now we convert the mass of NH3 into moles:
18.0 g NH3 × (1 mol NH3 / 17.03 g NH3) = 1.057 moles NH3
According to the balanced equation, 3 moles of NH3 produce 1 mole of N2. Therefore, the moles of N2 produced will be:
1.057 moles NH3 × (1 mol N2/3 moles NH3) = 0.352 moles N2
Last, we convert the moles of N2 to grams:
0.352 moles N2 × (28.02 g N2/mol) = 9.86 grams N2
Therefore, 9.86 grams of nitrogen gas will be formed.