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Virty [35]
3 years ago
9

11. How many grams of O2 are required to produce 36.9 grams of ZnO? 22n +102 → 2ZnO

Chemistry
1 answer:
Fittoniya [83]3 years ago
3 0

Answer:

Mass = 7.2 g

Explanation:

Given data:

Mass of ZnO = 36.9 g

Mass of oxygen needed = ?

Solution:

Chemical equation:

2Zn + O₂ → 2ZnO

Number of moles of ZnO:

Number of moles = Mass/ molar mass

Number of moles = 36.9 g/81.38 g/mol

Number of moles = 0.45 mol

Now we will compare the moles of ZnO with oxygen:

                      ZnO           :           O₂

                         2             :            1

                      0.45           :       1/2 × 0.45 = 0.225

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.225 mol × 32 g/mol

Mass = 7.2 g

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Answer : The percent yield is, 83.51 %

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First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

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