OpenStudy (anonymous):

If 4.88 grams of Zn reacts with 5.03 grams of S8 to produce 6.02 grams of ZnS, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem. unbalanced equation: Zn + S8 yields ZnS Someone please help me!!!

3 years ago
OpenStudy (jfraser):

first and most important, balance the equation

3 years ago
OpenStudy (joannablackwelder):

Yep, @JFraser is right. Do you know how to balance, @firesquad ?

3 years ago
OpenStudy (anonymous):

Balanced equation: 8Zn + S8 ---> 8ZnS

3 years ago
OpenStudy (anonymous):

@JFraser @JoannaBlackwelder

3 years ago
OpenStudy (joannablackwelder):

Yep, good. Percent yield = (actual/theoretical)*100

3 years ago
OpenStudy (joannablackwelder):

Actual is the amount of product given in the problem.

3 years ago
OpenStudy (joannablackwelder):

Theoretical is the amount of product found using stoichiometry from the amount of reactant given.

3 years ago
OpenStudy (joannablackwelder):

*from the amount of limiting reactant

3 years ago
OpenStudy (anonymous):

I am stuck right here 4.88 g Zn x (1 mol Zn/ 65.38 g Zn) = 0.07464 mol Zn 5.03 g S8 x (1 mol S8/ 256.48 g S8) = 0.01961 mol S8

3 years ago
OpenStudy (joannablackwelder):

Ok, now use the molar ratios from the balanced reaction to convert to moles of ZnS

3 years ago
OpenStudy (anonymous):

what do i do after

3 years ago
OpenStudy (joannablackwelder):

For both calculations.

3 years ago
OpenStudy (anonymous):

would the ratio be 1: 8

3 years ago
OpenStudy (joannablackwelder):

For S8:ZnS, yes

3 years ago
OpenStudy (joannablackwelder):

But not for Zn:ZnS

3 years ago
OpenStudy (anonymous):

0.07464 mol Zn * (1/1)= 0.07464 mol Zn 0.01961 mol S8 * (8/1) = 0.1569 mol S8 is that how to do it

3 years ago
OpenStudy (joannablackwelder):

Well, that looks good for everything but the units at the end.

3 years ago
OpenStudy (anonymous):

What do I have to change for the units?

3 years ago
OpenStudy (joannablackwelder):

0.07464 mol ZnS 0.1569 mol ZnS

3 years ago
OpenStudy (joannablackwelder):

See why I did that?

3 years ago
OpenStudy (anonymous):

Oh Ok, now I find the percent yield right?

3 years ago
OpenStudy (joannablackwelder):

Well, we need to first determine which value of product to go with.

3 years ago
OpenStudy (joannablackwelder):

This is the limiting reactant part. Do you know how to determine that?

3 years ago
OpenStudy (anonymous):

The limiting reactant is S8

3 years ago
OpenStudy (joannablackwelder):

No, the limiting reactant is the reactant that produces less product.

3 years ago
OpenStudy (anonymous):

Sorry, it is Zn

3 years ago
OpenStudy (joannablackwelder):

Right :-)

3 years ago
OpenStudy (joannablackwelder):

So, what is the mass of product using the limiting reactant?

3 years ago
OpenStudy (anonymous):

Sorry for wait I had to do something. It is 0.07464 mol ZnS

3 years ago
OpenStudy (joannablackwelder):

No worries. That is the moles of product. What is the mass of product?

3 years ago
OpenStudy (anonymous):

65.38 g Zn

3 years ago
OpenStudy (joannablackwelder):

Hm, that's not what I get. How did you do that?

3 years ago
OpenStudy (anonymous):

I got it from the first formula

3 years ago
OpenStudy (joannablackwelder):

Hm, let me show you how I would do this step: 0.07464 mol ZnS(97.44 g/1 mol) = ? g ZnS

3 years ago
OpenStudy (anonymous):

Ok I get it

3 years ago
OpenStudy (joannablackwelder):

:-) So, can you find the percent yield?

3 years ago
OpenStudy (anonymous):

Wait 7.27 is the theoretical yield

3 years ago
OpenStudy (joannablackwelder):

Yep

3 years ago
OpenStudy (anonymous):

so what is the actual yield?

3 years ago
OpenStudy (joannablackwelder):

The amount of product given in the problem.

3 years ago
OpenStudy (anonymous):

Thank you so much

3 years ago
OpenStudy (joannablackwelder):

You're welcome! :-)

3 years ago