Question

) determine the theoretical yield and the percent yield if 21.8 g of k2co3 is produced from reacting 27.9 g ko2 with 29.0 l of co2 (at stp). the molar mass of ko2 = 71.10 g/mol and k2co3 = 138.21 g/mol.

Answer 1

The Balanced Chemical Equation is as follow;

                         4 KO₂  +  2 CO₂    →    2 K₂CO₃  +  3 O₂

First find out the Limiting Reagent,
According to equation,

         284 g (4 moles) KO₂ reacted with  =  44.8 L (2 moles) of CO₂
So,
                  27.9 g of KO₂ will react with  =  X  L of CO₂


Solving for X,
                          X  =  (44.8 L × 27.9 g) ÷ 284 g

                          X  =  4.40 L of CO₂

Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,

According to eq.

         284 g (4 moles) KO₂ formed  =  138.2 g of K₂CO₃
So,
         27.9 g of KO₂ will form  =  X g of K₂CO₃

Solving for X,
                        X  =  (138.2 g × 27.9 g) ÷ 284 g

                        X  =  13.57 g of K₂CO₃

So, 13.57 g of K₂CO₃ formed is the theoretical yield.

%age Yield  =  13.57 / 21.8 × 100

%age Yield  =  62.24 %

Answer 2

Answer : The theoretical yield of $K_2CO_3$ = 27.089 g

The percent yield of $K_2CO_3$  is, 80.47 %

Explanation :  Given,

Mass of $KO_2$ = 27.9 g

Volume of $CO_2$ = 29.0 L  (At STP)

Molar mass of $KO_2$ = 71.10 g/mole

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $K_2CO_3$ = 138.21 g/mole

First we have to calculate the moles of $CO_2$ and $KO_2$.

At STP,

As, 22.4 L volume of $CO_2$ present in 1 mole of $CO_2$

So, 29.0 L volume of $CO_2$ present in $\frac{29.0}{22.4}=1.29$ mole of $CO_2$

$\text{Moles of }KO_2=\frac{\text{Mass of }KO_2}{\text{Molar mass of }KO_2}=\frac{27.9g}{71.10g/mole}=0.392mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 4 moles of $KO_2$ react with 2 mole of $CO_2$

So, 0.392 moles of $KO_2$ react with $\frac{2}{4}\times 0.392=0.196$ 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 $K_2CO_3$.

As, 4 moles of $KO_2$ react to give 2 moles of $K_2CO_3$

So, 0.392 moles of $KO_2$ react to give $\frac{2}{4}\times 0.392=0.196$ moles of $K_2CO_3$

Now we have to calculate the mass of $K_2CO_3$.

$\text{Mass of }K_2CO_3=\text{Moles of }K_2CO_3\times \text{Molar mass of }K_2CO_3$

$\text{Mass of }K_2CO_3=(0.196mole)\times (138.21g/mole)=27.089g$

The theoretical yield of $K_2CO_3$ = 27.089 g

The actual yield of $K_2CO_3$ = 21.8 g

Now we have to calculate the percent yield of $K_2CO_3$

$\%\text{ yield of }K_2CO_3=\frac{\text{Actual yield of }K_2CO_3}{\text{Theoretical yield of }K_2CO_3}\times 100=\frac{21.8g}{27.089g}\times 100=80.47\%$

Therefore, the percent yield of $K_2CO_3$  is, 80.47 %