3cu(s) + 8 hno3(aq) → 3 cu(no3)2(aq) + 2 no(g) + 4 h20(1) a) if 5.58 g of copper(ii) nitrate, cu(no3)2, is eventually obtained, how many grams of nitrogen monoxide, no, would have formed?

The mass of nitrogen monoxide formed is 0.6 grams.

Explanation:

To calculate the number of moles, we use the equation:

   .....(1)

Given mass of copper (II) nitrate = 5.58 g

Molar mass of copper (II) nitrate = 187.56 g/mol

Putting values in above equation, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

When 3 moles of copper (II) nitrate is formed, then 2 moles of nitrogen monoxide is formed.

So, when 0.03 moles of copper (II) nitrate is formed, then of nitrogen monoxide is formed.

Now, calculating the mass of nitrogen monoxide by using equation 1, we get:

Molar mass of nitrogen monoxide = 30 g/mol

Moles of nitrogen monoxide = 0.02 moles

Putting values in equation 1, we get:

Hence, the mass of nitrogen monoxide formed is 0.6 grams.

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.