39.7k views
0 votes
what is the percent yield of NaCl if 31.0 g of CuCl reacts with excess NaNo3 to produce 21.2 g of NaCl

User Melwil
by
7.1k points

1 Answer

5 votes
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.

Balanced Equation: CuCl + NaNO₃ → NaCl + CuNO₃

Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol

the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,

then moles of NaCl = 0.31 mol

Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g

⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.

Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %


NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.
User Juliusz
by
8.6k points