Question

Consider the following reaction:

Copper (II) chloride (mm = 135.45g/mol) reacts with sodium nitrate (mm=84.99 g/mol) to form sodium chloride (mm = 58.44 g/mol) and copper (II) nitrate (mm = 187.56 g/mol).

CuCl₂ (aq) + 2NaNO₃ (aq) → 2NaCl (aq) + Cu(NO₃)₂ (aq)

Using your balanced Chemical Reaction from the previous question and that 15.3 grams of copper (II) chloride are reacted with 20.5 grams of sodium nitrate.

If the actual yield of the reaction is 11.3 grams of sodium chloride, what is the percent yield for the reaction?

Answer 1

The percent yield for the reaction is 85.80%.

Step-by-step explanation:

To calculate the percent yield for this reaction, we need to first calculate the theoretical yield of sodium chloride. We can do this by converting the given mass of copper (II) chloride to moles using the molar mass of copper (II) chloride, and then using the stoichiometric ratio from the balanced equation to calculate the moles of sodium chloride. Finally, we can convert the moles of sodium chloride back to grams using the molar mass of sodium chloride.

Given:

Mass of copper (II) chloride = 15.3 g

Mass of sodium nitrate = 20.5 g

Actual yield of sodium chloride = 11.3 g

Using the molar mass of copper (II) chloride (135.45 g/mol), moles of copper (II) chloride can be calculated as:

Moles of CuCl₂ = mass of CuCl₂ / molar mass of CuCl₂

Moles of CuCl₂ = 15.3 g / 135.45 g/mol = 0.113 mol CuCl₂

Using the stoichiometric ratio from the balanced equation (1:2), moles of sodium chloride can be calculated as:

Moles of NaCl = 2 × moles of CuCl₂ = 2 × 0.113 mol = 0.226 mol NaCl

Finally, we can calculate the theoretical yield of sodium chloride in grams:

Theoretical yield = moles of NaCl × molar mass of NaCl

Theoretical yield = 0.226 mol × 58.44 g/mol = 13.18 g

Now we can calculate the percent yield:

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

Percent yield = (11.3 g / 13.18 g) × 100

Percent yield = 85.80%