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.

Answers: 1

Show answers

Answers

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 %

Answer from: Quest

What’s the question

Answer from: Quest

You can still be friends with each other. you don't have to date since he likes someone else. just stick to being friends for now

Another question on Chemistry

Chemistry, 22.06.2019 09:20

Sugar is dissolved in water. which is the solute? sugar neither both water

Answers: 1

Answer

Chemistry, 22.06.2019 17:00

The atoms of a solid aluminum can are close together, vibrating in a rigid structure. if the can is warmed up on a hot plate, what happens to the atoms?

Answers: 3

Answer

Chemistry, 22.06.2019 20:00

What is the molarity of the solution produced when 145 g of nacl is dissolved in sufficient water to prepare 2.75 l of solution?

Answers: 1

Answer

Chemistry, 22.06.2019 21:50

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes

Answers: 1

Answer

You know the right answer?

)determine the theoretical yield and the percent yield if 21.8 g of k2co3 is produced from reacting...

Questions