Question

When 15 grams of copper (II) chloride (CuCl2) reacts with 20 grams of sodium nitrate (NaNO3), 11.3 grams of sodium chloride (NaCl) is obtained. What is the percent yield of this reaction?

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

Answer 1

Proportionally, for every one mole of sodium nitrate, you should yield one mole of sodium chloride.
Sodium nitrate's molar mass is 85.0 grams/mole:
20 grams/(85.0 grams/mole) = 0.24 moles of sodium nitrate
The theoretical yield of sodium chloride would also be 0.24 moles
The molar mass of sodium chloride is 58.5 grams/mole:
11.3 grams/(58.5 grams/mole) = 0.19 moles
Percent yield = ((actual yield)/(theoretical yield)) * 100%
Percent yield = (0.19 moles/0.24 moles) * 100% = 79% yield

Answer 2

Answer : The percent yield of the reaction is, 86.3 %

Solution : Given,

Mass of
`CuCl_2` = 15 g

Mass of
`NaNO_3` = 20 g

Molar mass of
`CuCl_2` = 134.45 g/mole

Molar mass of
`NaNO_3` = 84.9 g/mole

Molar mass of
`NaCl` = 58.44 g/mole

First we have to calculate the moles of
`CuCl_2` and
`NaNO_3`.

`\text{ Moles of }CuCl_2=\frac{\text{ Mass of }CuCl_2}{\text{ Molar mass of }CuCl_2}=(15g)/(134.45g/mole)=0.112moles`

`\text{ Moles of }NaNO_3=\frac{\text{ Mass of }NaNO_3}{\text{ Molar mass of }NaNO_3}=(20g)/(84.9g/mole)=0.236moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2`

From the balanced reaction we conclude that

As, 1 mole of
`CuCl_2` react with 2 mole of
`NaNO_3`

So, 0.112 moles of
`CuCl_2` react with
`0.112* 2=0.224` moles of
`NaNO_3`

From this we conclude that,
`NaNO_3` is an excess reagent because the given moles are greater than the required moles and
`CuCl_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`NaCl`

From the reaction, we conclude that

As, 1 mole of
`CuCl_2` react to give 2 mole of
`NaCl`

So, 0.112 moles of
`CuCl_2` react to give
`0.112* 2=0.224` moles of
`NaCl`

Now we have to calculate the mass of
`NaCl`

`\text{ Mass of }NaCl=\text{ Moles of }NaCl* \text{ Molar mass of }NaCl`

`\text{ Mass of }NaCl=(0.224moles)* (58.44g/mole)=13.09g`

Theoretical yield of
`NaCl` = 13.09 g

Experimental yield of
`NaCl` = 11.3 g

Now we have to calculate the percent yield of
`NaCl`

`\% \text{ yield of }NaCl=\frac{\text{ Experimental yield of }NaCl}{\text{ Theretical yield of }NaCl}* 100`

`\% \text{ yield of }NaCl=(11.3g)/(13.09g)* 100=86.3\%`

Therefore, the percent yield of the reaction is, 86.3 %