Question

In the reaction, 145.93 grams 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 moles?

Answer 1

In this question, we have the following reaction:

2 Cu(NO3)2 -> 2 CuO + 4 NO2 + O2, this is the reaction of decomposition of Copper II nitrate, and it is balanced.

We have 145.93 grams of Cu(NO3)2, the molar mass of this compound is 187.56g/mol, now that we have the given mass and the molar mass, we can find the number of moles of copper II nitrate:

187.56g = 1 mol of Cu(NO3)2

145.93g = x moles of Cu(NO3)2

187.56x = 145.93

x = 145.93/187.56

x = 0.778 moles of Cu(NO3)2 in 145.93 grams of it

Now that we have the number of moles of this compound, we can find the number of moles of any substance on the products side, we need to follow the molar ratios in the question, which are:

2 Cu(NO3)2 = 2 CuO

2 Cu(NO3)2 = 4 NO2

2 Cu(NO3)2 = 1 O2

This is how much it takes of Cu(NO3)2, proportionally, to produce each product.

For CuO, we will have:

2 Cu(NO3)2 = 2 CuO

0.778 Cu(NO3)2 = x CuO

x = 0.778 moles of CuO

For NO2:

2 Cu(NO3)2 = 4 NO2

0.778 Cu(NO3)2 = x NO2

2x = 3.112

x = 3.112/2

x = 1.556 moles of NO2

For O2:

2 Cu(NO3)2 = 1 O2

0.778 Cu(NO3)2 = x O2

2x = 0.778

x = 0.778/2

x = 0.389 moles of O2

Therefore the number of moles for each of the products, based on 0.778 moles of Cu(NO3)2 is:

0.778 moles of CuO

1.556 moles of NO2

0.389 moles of O2