To solve this, we need to use stoichiometry. From the balanced equation:
1 mol CuSO4 + 4 mol NH3 → 1 mol Cu(NH3)4SO4
From this, we can see that for every 1 mol of CuSO4, we need 4 mol of NH3 to produce 1 mol of Cu(NH3)4SO4.
We can use the molar mass of NH3 to convert 6g to moles:
6g NH3 ÷ 17.03 g/mol = 0.352 mol NH3
Using the ratio from the balanced equation, we can then determine the theoretical yield of Cu(NH3)4SO4:
0.352 mol NH3 ÷ 4 = 0.088 mol Cu(NH3)4SO4
To convert to mass, we can use the molar mass of Cu(NH3)4SO4:
0.088 mol Cu(NH3)4SO4 x 245.7 g/mol = 21.7 g Cu(NH3)4SO4
Therefore, theoretically, 21.7 g of Cu(NH3)4SO4 could be produced.
b. The percent yield can be calculated using the actual yield and theoretical yield. We were given the actual yield of Cu(NH3)4SO4, which is 12.6 g.
The percent yield formula is:
Percent yield = (actual yield ÷ theoretical yield) x 100%
Substituting the values we have:
Percent yield = (12.6 g ÷ 21.7 g) x 100% = 58%
Therefore, the percent yield of Cu(NH3)4SO4 is 58%.