Question

15 grams of copper (II) chloride react with 20 grams of sodium nitrate If 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction? CuCl2+2NaNO3=Cu(NO3)2+2NaCl

Answer 1

The percent yield for the reaction = 81.88%

Step-by-step explanation

Given:

Mass of CuCl2 = 15 grams

Mass of NaNO3 = 20 grams

Actual yield of NaCl = 11.3 grams

Equation: CuCl₂ + 2NaNO₃ ----> Cu(NO₃)₂ + 2NaCl

What to find:

The percent yield for the reaction.

Step-by-step solution:

Step 1: Covert the given grams to moles.

Molar mass of CaCl₂ = 110.98 g/mol

Molar mass of NaNO₃ = 84.995 g/mol

The mass of the reactants can be converted to moles using the mole formula

`Mole=\frac{Mass}{Molar\text{ }mass}`

Moles of CaCl₂

`Mole=\frac{15g}{110.98\text{ }g\text{/}mol}=0.1352\text{ }mol\text{ }CaCl_2`

Moles of NaNO₃ is

`Mole=\frac{20\text{ }g}{84.995\text{ }g\text{/}mol}=0.2353\text{ }mol\text{ }NaNO₃`

Step 2: Determine the limiting reactant.

From the equation, 1 mole CaCl₂ required 2 moles NaNO₃

So 0.1352 moles CaCl₂ will require (0.1352 x 2)/1 = 0.2704 moles NaNO₃

Note that the moles present in 20 grams NaNO₃ is 0.2353 mol, but 0.2704 mol NaNO₃ is needed to consume all the 0.1352 mol CaCl₂. This implies NaNO₃ is the limiting reactant and CaCl₂ is the reactant in excess.

Step 3: Calculate the theoretical yield for the reaction.

From the equation, 2 moles NaNO₃ produce 2 moles NaCl

Therefore 0.2353 moles NaNO₃ will produce

`\frac{0.2353mol\text{ }NaNO_3*2mol\text{ }NaCl}{2mol\text{ }NaNO_3}=0.2353\text{ }mol\text{ }NaCl_`

Step 4: Convert 0.2353 mol NaCl to mass.

Molar mass of NaCl = 58.44 g/mol

So 0.2353 mol = Mass/58.44 g/mol

Mass = 0.2353 mol x 58.44 g/mol

Mass of NaCl = 13.8 grams

Step 3: Determine the percent yield of the reaction.

The percent yield of the reaction can be calculated using the formula

`Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}*100\%`

Actual yield of NaCl = 11.3 grams

Theoretical yield of NaCl = 13.8 grams

Thus,

`Percent\text{ }yield=\frac{11.3\text{ }grams}{13.8\text{ }grams}*100\%=81.88\%`

The percent yield for the reaction = 81.88%