Question

In the reaction, 64.92 moles of copper (II) nitrate react, to give, as products copper (II) oxide and nitrogen dioxide and molecular oxygen. How much product is produced of copper (II) oxide, in moles, nitrogen dioxide, in moles, and molecular oxygen, in grams?

Show all work, use unit-cancellation method is necessary, show super- and subscripts.

Answer 1

Given:

64.92 moles of copper (II) nitrate (Cu(NO3)2)

Reactants:

Cu(NO3)2 (copper (II) nitrate)

Products:

CuO (copper (II) oxide)

NO2 (nitrogen dioxide)

O2 (molecular oxygen)

Molar Mass:

Cu(NO3)2 (copper (II) nitrate) = 187.56 g/mol

CuO (copper (II) oxide) = 79.55 g/mol

NO2 (nitrogen dioxide) = 46.01 g/mol

O2 (molecular oxygen) = 32.00 g/mol

Step 1: Calculate moles of CuO produced:

Using the balanced chemical equation:

2 Cu(NO3)2 → 2 CuO + 4 NO2 + O2

1 mole of Cu(NO3)2 produces 2 moles of CuO

64.92 moles of Cu(NO3)2 * (2 moles CuO / 1 mole Cu(NO3)2) = 129.84 moles of CuO

Step 2: Calculate moles of NO2 produced:

Using the balanced chemical equation:

2 Cu(NO3)2 → 2 CuO + 4 NO2 + O2

1 mole of Cu(NO3)2 produces 4 moles of NO2

64.92 moles of Cu(NO3)2 * (4 moles NO2 / 1 mole Cu(NO3)2) = 259.68 moles of NO2

Step 3: Calculate grams of O2 produced:

Using the balanced chemical equation:

2 Cu(NO3)2 → 2 CuO + 4 NO2 + O2

1 mole of Cu(NO3)2 produces 1 mole of O2

64.92 moles of Cu(NO3)2 * (1 mole O2 / 1 mole Cu(NO3)2) = 64.92 moles of O2

Convert moles of O2 to grams:

64.92 moles of O2 * 32.00 g/mol = 2077.44 grams of O2

Results:

The amount of CuO produced is 129.84 moles

The amount of NO2 produced is 259.68 moles

The amount of O2 produced is 2077.44 grams