QUESTION 4 If 278 mg of calcium nitrate is reacted with 257 mg of potassium phos

Question

QUESTION 4 If 278 mg of calcium nitrate is reacted with 257 mg of potassium phosphate, caldum phosphate and potassium nitrate products will be obtained according to the following reaction 3Ca(NO312 + 2 K3PO4 Ca3(PO4)2 + 6 KNO3 What is the theoretical yield of potassium nitrate in milligrams? (Do not try to include units in your answer as BB will not understand it.) QUESTION S is reacted with 257 mg of potassium phosphate, calcium phosphate and potassium nitrate products wll be reaction: If 207 mg of potassium nitrate was made in reaction, what was the percent yield of the reaction? (Do not

Explanation / Answer

3Ca(NO3)2 + 2K3PO4------------> Ca3(PO4)2 + 6KNO3

no of moles of Ca(NO3)2   = W/G.M.Wt

                                           = 278*10^-3/164   = 0.0017 moles

no of moles of K3PO4   = W/G.M.Wt

                                       = 257*10^-3/212.27 =0.0012 moles

2 moles of K3PO4 react with 3 moles of Ca(NO3)2

0.0012 moles of K3PO4 react with = 3*0.0012/2   = 0.0018 moles of Ca(NO3)2

Ca(NO3)2 is limiting reactant

3 moles of Ca(NO3)2 react with K3PO4 to gives 6 moles of KNO3

0.0017 moles of Ca(NO3)2 react with K3Po4 to gives = 6*0.0017/3   = 0.0034 moles of KNO3

mass of KNO3 = no of moles * gram molar mass

                           = 0.0034*101   = 0.3434g   = 343.4mg

The theoritical yield of KNO3 = 343.4mg

actual yield of KNO3              = 207mg

percentage yield   = actual yield*100/theoritical yield

                             = 207*100/343.4   = 60.23%

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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