Question

The deep blue compound Cu(NH3)4SO4 is made by the reaction of copper(ll) sulfate and ammonia. CuSO4 (aq) + 4 NH3(aq) + Cu(NH3)4SO4 (aq) If you use 36.0 g of CuSO4 and excess NH3, what is the theoretical yield of Cu(NH3)4SO4? ________g Cu(NH3)SO4 If you isolate 30.2 g of Cu(NH3)SO4, what is the percent yield of Cu(NH3)4 SO4?______%

Answer 1

The theoretical yield of Cu(NH3)4SO4 is found to be 55.40 g using stoichiometry, and the percent yield is calculated as 54.51% using the actual isolated amount of 30.2 g of the compound.

Step-by-step explanation:

To calculate the theoretical yield of Cu(NH3)4SO4, we need to convert the mass of CuSO4 to moles by using its molar mass (159.62 g/mol). For every mole of CuSO4, one mole of Cu(NH3)4SO4 is produced, according to the stoichiometry of the reaction. Therefore, the theoretical yield in grams is found using the molar mass of Cu(NH3)4SO4 (245.79 g/mol). To find the percent yield, you use the formula (actual yield / theoretical yield) × 100%. In this case, the percent yield is calculated using the isolated 30.2 g of Cu(NH3)4SO4.

The theoretical yield calculation is as follows:

• Calculate moles of CuSO4: 36.0 g ÷ 159.62 g/mol = 0.2255 mol.
• Calculate moles of Cu(NH3)4SO4: 1:1 ratio, so also 0.2255 mol.
• Calculate mass of Cu(NH3)4SO4: 0.2255 mol × 245.79 g/mol = 55.40 g (theoretical yield).

The percent yield calculation is as follows:

Percent yield = (30.2 g / 55.40 g) × 100% = 54.51%.