Limiting reactant For the following balanced chemical reaction, determine the fo

Question

Limiting reactant For the following balanced chemical reaction, determine the following: 2(NH_4)_3, PO_4(aq) + 3Cu(C_2H_3O_2)(aq) rightarrow Cu_3(PO_4)_2(s) + 6NH_4C_2H_3O_2(qq) a. Determine the limiting reactant and theoretical yield (in mass of product) for the following combinations. i. 35.0 g each of both reactants. ii. 45.0 g of ammonium phosphate and 35.0 g of copper(II) acetate. iii. 35.0 g of ammonium phosphate and 45.0 g of copper(II) acetate. iv. 10.05 g of ammonium phosphate and 20.10 g of copper(II) acetate. v. 20.10 g of ammonium phosphate and 10.05 g of copper(II) acetate. vi. 35.0 mL each of a 1.35 M solution of both reactants. vii. 45.0 ml of a 1.67 M of ammonium phosphate and 13.5 ml of a 5.67 M solution of copper(II) acetate. viii. 78.0 mL of a 0.897 M solution of copper(II) acetate and 90.78 ml of a 1.45 M solution of ammonium phosphate. ix. 125 mL of a 0.345 M solution of ammonium phosphate and 56.0 g of copper (II) acetate. x. 156 mL of a 0.176 M solution of copper(II) acetate and 0.5607 g of ammonium phosphate. For each of the calculations in part (a), determine the percent yield of the reactant if 1.780 g of solid product is obtained. For each of the percent yields determine in part (b), comment on the validity of the experiment. b. Using only valid percent yields from part(b), determine the mass of solid product excerpted if 1235 g of each reactant is used.

Explanation / Answer

General calculations that willbe required:

Molecular mass of reactant A=149g

Molecular mass of reactant B=181g

Molecular mass of product C= 380g

A) i) for 35g of A

according to the balanced chemical equation:

298g of A ---> 380g C

35g A...> 380/298*35= 44g of C

to get 44g C amount of B required = 544/380*44= 62g B

but the present of of B is only 35g , Hence B is the limiting agent...

therefore.. from 35g of B the amount of C obtained =380/544*35=24.4g of C

ii) similar as above: for 45 g of A the amount of produt C =380/298*45=57.4g of C

the amount of B required to make 57.4g of C= 544/380*57.4=82g of B

but since we have only 35g of B .therefore B is the limiting agent

Hence the amount of product C formed from 35g of B= 380/544*35= 24.4g of C

iii) similar as above , for 35 g of A the amount of product C formed = 380/298*35=44.6g of C

for 44.6 g of C the amount of B required is=544/380*44.6=63.8g of B

since amount of B present is only 45 g hence B is the limiting agent

therefore the amount of C formed from 45g of B =380/544*45= 31.43g of C

iv) Similar as above , for 20.1 g of B the amount of C formed is=380/544*20.1=13.9g of C

to get 13.9 g of C , the amount of A required is= 298/380*13.9= 10.9g of A

but the amount of A present is 10.05g, hence A is the limiting agent.

therefore, the amount of C formed from 10.05g of A=380/298*10.05=13.3g of C

B) i) %yeild= 1.78/24.2*100= 7.2% ii) %yeild=1.78/24.4*100= 7.2%

   iii) %yeild= 1.78/31.43*100= 5.6% iv) %yeild= 1.78/13.3*100= 13.3%

C) The last option i.e 10g A and 20g B is most valid as we are getting some substantial amount of product for the same, whereas in other cases the % yeild is less than even 10 % which can be ignored.

D) from 298g of A amount of C obtained = 380g

from 1235g of A the amount of C obtained= 380/298*1235= 1574g of C

now to get 380g of C the amount of B required is = 544g

to get 1574g of C = 544/380*1574= 2253g of B is required... but we hve only 1235g of B , hence B is the limitng agent

therefore

since.. 544g of B gives 380g of C , therefore

1235g of B will give= 380/544*1235= 862g of C

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.