Consider the reaction Cu3PO4+Fe(NO3)2->CuNO3+Fe3(PO4)2 1. Balance the equation 2

ID: 869649 • Letter: C

Question

Consider the reaction Cu3PO4+Fe(NO3)2->CuNO3+Fe3(PO4)2

1. Balance the equation

2. Determine the molar masses

3. How many moles of copper phosphate reacts with 5 moles of iron nitrate?

4. How many grams of each of the compounds above are reacted?

5. Using your answers in question 4 how many grams of copper nitrate are formed from copper phosphate and iron nitrate?

6. What is your limiting reactant?

7. What is your theoretical yield?

8. If your actual yield is 1242.45 grams what is your percent yield?

1) Balanced equation:
2 Cu3PO4 + 3 Fe(NO3)2 ------------------->6 CuNO3 + Fe3(PO4)2

2) molar masses Cu3PO4 = 285.60 g/mol

Fe(NO3)2 = 179.8g/mol

CuNO3 =125.55 g/mol

Fe3(PO4)2 = 357.47g/mol

3) 2 moles of copper phosphate reacts with 3 moles of iron nitrate

how many moles of copper phosphate reacts with 5 moles of iron nitrate ?

moles of copper phosphate = 2x 5/3

= 3.33

4) 2 Cu3PO4 + 3 Fe(NO3)2 ------------------->6 CuNO3 + Fe3(PO4)2

2 x 285.6 3 x 179.8 6 x 125.55 357.47

571.2 g 539.4 g 753.3g 357.47 g (each reacted compounds in gm)

5) 753.3g CuNO3 formed

6) Fe(NO3)2 weight is less . so it is limiting reagent

7) theoretical yield = 753.3 + 357.47= 1110 .77g

8) percent yield = (actual yield /theoretical yield) x100

= 1242.45 x100/1110 .77

= 111.8%

check once problem (8) actual yield value

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