Question

Excess copper reacts with 170. g of silver nitrate to produce copper (II) nitrate and silver. Determine the % yield of Cu(NO3)2 if the actual yield is 46.9 g.

Answer 1

The percent yield of Cu(NO3)2 can be calculated using the formula: Percent yield = (actual yield / theoretical yield) x 100. In this case, the actual yield is 46.9 g. To determine the theoretical yield, we need to calculate the moles of copper (II) nitrate that could be formed.

Step-by-step explanation:

The percent yield of Cu(NO3)2 can be calculated using the formula:

Percent yield = (actual yield / theoretical yield) x 100

In this case, the actual yield is given as 46.9 g. To determine the theoretical yield, we need to calculate the moles of copper (II) nitrate that could be formed. From the balanced chemical equation:

Cu (s) + 2 AgNO3 (aq) → Cu(NO3)2 (aq) + 2 Ag (s)

We can see that the mole ratio between copper (II) nitrate and silver nitrate is 1:2. Therefore, the moles of copper (II) nitrate are calculated as:

moles Cu(NO3)2 = (moles AgNO3 / 2)

Using the molar mass of AgNO3 (169.88 g/mol):

moles AgNO3 = (mass AgNO3 / molar mass AgNO3)

moles AgNO3 = (170 g / 169.88 g/mol) = 1 mol

Therefore, moles Cu(NO3)2 = (1 mol / 2) = 0.5 mol

The molar mass of Cu(NO3)2 is 187.57 g/mol. Therefore, the theoretical yield of Cu(NO3)2 is:

theoretical yield = (moles Cu(NO3)2 x molar mass Cu(NO3)2)

theoretical yield = (0.5 mol x 187.57 g/mol) = 93.79 g

Now, we can calculate the percent yield:

percent yield = (46.9 g / 93.79 g) x 100

percent yield = 50%