Question

Determine the theoretical yield and the percent yield if 21.8 g of KCO is produced from reacting 27.9 g KO with 29.0 L of CO (at STP). The molar mass of KO = 71.10 g/mol and KCO = 138.21 g/mol. 4 KO(s) + 2 CO(g) → 2 KCO(s) + 3 O(g)

Answer 1

Answer : The theoretical yield and the percent yield is, 27.1 grams and 80.4 % respectively.

Step-by-step explanation:

First we have to calculate the moles of KO₂ and CO₂.

`\text{Moles of }KO_2=\frac{\text{Mass of }KO_2}{\text{Molar mass of }KO_2}`

`\text{Moles of }KO_2=(27.9g)/(71.10g/mol)=0.392mol`

and,

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 22.4 L volume of CO₂ present in 1 mole of CO₂ gas

So, 29.0 L volume of CO₂ present in
`(29.0)/(22.4)=1.29` mole of CO₂ gas.

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2`

From the balanced reaction we conclude that

As, 4 mole of
`KO_2` react with 2 mole of
`CO_2`

So, 0.392 moles of
`KO_2` react with
`(2)/(4)* 0.392=0.196` moles of
`CO_2`

From this we conclude that,
`CO_2` is an excess reagent because the given moles are greater than the required moles and
`KO_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`K_2CO_3`

From the reaction, we conclude that

As, 4 mole of
`KO_2` react to give 2 mole of
`K_2CO_3`

So, 0.392 moles of
`KO_2` react to give
`(2)/(4)* 0.392=0.196` moles of
`K_2CO_3`

Now we have to calculate the mass of
`K_2CO_3`

`\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3* \text{ Molar mass of }K_2CO_3`

Molar mass of
`K_2CO_3` = 110.98 g/mole

`\text{ Mass of }K_2CO_3=(0.196moles)* (138.21g/mole)=27.1g`

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

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

Experimental yield = 21.8 g

Theoretical yield = 27.1 g

Now put all the given values in this formula, we get:

`\text{Percent yield}=(21.8g)/(27.1g)* 100=80.4\%`

Therefore, the percent yield of the reaction is, 80.4 %