Chemistry, 07.04.2020 22:58 timjape3g3z

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)

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Answer from: zitterkoph

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

Explanation:

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

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

$\text{Moles of }KO_2=\frac{27.9g}{71.10g/mol}=0.392mol$

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 $\frac{29.0}{22.4}=1.29$ 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 $\frac{2}{4}\times 0.392=0.196$ 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

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 $\frac{2}{4}\times 0.392=0.196$ moles of K_2CO_3

Now we have to calculate the mass of K_2CO_3

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

Molar mass of K_2CO_3 = 110.98 g/mole

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

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

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 21.8 g

Theoretical yield = 27.1 g

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

$\text{Percent yield}=\frac{21.8g}{27.1g}\times 100=80.4\%$

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

Answer from: Quest

Do u have the n? or maybe the m?

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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