Chemistry, 20.03.2020 09:51, grangian06

Determine the theoretical yield of ZnS that can be prepared from the reaction of 0.488 g Zn with 0.503 g S8. What is the percent yield, if 0.606 g of ZnS were experimentally recovered

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answered: Webber07

The percentage yield of zinc sulfide is 82.90 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For Zinc:

Given mass of zinc = 0.488 g

Molar mass of zinc = 65.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of zinc}=\frac{0.488g}{65.4g/mol}=0.0075mol$

For sulfur:

Given mass of sulfur = 0.503 g

Molar mass of sulfur = 256 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfur}=\frac{0.503g}{256g/mol}=0.00196mol$

The given chemical equation follows:

$8Zn+S_8\rightarrow 8ZnS$

By Stoichiometry of the reaction:

8 moles of zinc reacts with 1 mole of sulfur

So, 0.0075 moles of zinc will react with = $\frac{1}{8}\times 0.0075=0.00094mol$ of sulfur

As, given amount of sulfur is more than the required amount. So, it is considered as an excess reagent.

Thus, zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

8 moles of zinc produces 8 moles of zinc sulfide

So, 0.0075 moles of zinc will produce = $\frac{8}{8}\times 0.0075=0.0075moles$ of zinc sulfide

Now, calculating the mass of zinc sulfide from equation 1, we get:

Molar mass of zinc sulfide = 97.5 g/mol

Moles of zinc sulfide = 0.0075 moles

Putting values in equation 1, we get:

$0.0075mol=\frac{\text{Mass of zinc sulfide}}{97.5g/mol}\\\\\text{Mass of zinc sulfide}=(0.0075mol\times 97.5g/mol)=0.731g$

To calculate the percentage yield of zinc sulfide, we use the equation:

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

Experimental yield of zinc sulfide = 0.606 g

Theoretical yield of zinc sulfide = 0.731 g

Putting values in above equation, we get:

$\%\text{ yield of zinc sulfide}=\frac{0.606g}{0.731g}\times 100\\\\\% \text{yield of zinc sulfide}=82.90\%$

Hence, the percentage yield of zinc sulfide is 82.90 %

answered: Guest

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answered: Guest

Hi; /c