Chemistry, 21.07.2019 23:00 asuhdude57

)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.

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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 from: dmgcity

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 %

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