Question

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of this combination reaction

Answer 1

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of
`Na` = 5.0 g

Mass of
`Cl_2` = 10.0 g

Molar mass of
`Na` = 23 g/mol

Molar mass of
`Cl_2` = 71 g/mol

First we have to calculate the moles of
`Na` and
`Cl_2`.

`\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}`

`\text{Moles of }Na=(5.0g)/(23g/mol)=0.217mol`

and,

`\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}`

`\text{Moles of }Cl_2=(10.0g)/(71g/mol)=0.141mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

`2Na+Cl_2\rightarrow 2NaCl`

From the balanced reaction we conclude that

As, 2 mole of
`Na` react with 1 mole of
`Cl_2`

So, 0.217 moles of
`Na` react with
`(0.217)/(2)=0.108` moles of
`Cl_2`

From this we conclude that,
`Cl_2` is an excess reagent because the given moles are greater than the required moles and
`Na` 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, 2 mole of
`Na` react to give 2 mole of
`NaCl`

So, 0.217 mole of
`HCl` react to give 0.217 mole of
`NaCl`

Now we have to calculate the mass of
`NaCl`

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

Molar mass of
`NaCl` = 58.5 g/mole

`\text{ Mass of }NaCl=(0.217moles)* (58.5g/mole)=12.7g`

Now we have to calculate the percent yield of this reaction.

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

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield =
`(5.24g)/(12.7g)* 100`

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %