QUESTION 4 If 278 mg of calcium nitrate is reacted with 257 mg of potassium phos
ID: 563614 • Letter: Q
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 notExplanation / 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%