Someone answered this and I figured the same but 1. Weight of beaker (g) 2. Weig

Question

Someone answered this and I figured the same but 1. Weight of beaker (g) 2. Weight of beaker and ferrous ammonium sulfate hexahydrate (g) 3. Weight of filter paper (g) 4. Weight of filter paper and product (g) 3.009 s. Weight of ferrous armon i um sulfate hexahydrate (g) 6. Moles of ferrous ammonium sul fate hexahydrate (mol) 7. Predicted number of moles of K,Fe(C,o.), 3H,o (mo1)5 8. Predicted number of grams of K,Fe (C O.3H20 (g) 9. Experimental weight (yield) ot K,pe (C, -3 3H,0 (g)305 10. Percentage yield of KaFe (C,04)3.3H20 (%) All steps in the calculation of the predicted weight of the product (Steps 6, 7, and 8): Same · Calculation of the percentage yield (Step 10): 9bydd: 30,k/00 3 -g x 100 100 Summary of observations (use the back of the report sheet Experiment #8 82

Explanation / Answer

Note that they are and should NOT be the same

Q7 --> moles of K3Fe(C2O4)3*3H2O

recall that moles are essentialy "amount of substance"

1 mol of K3Fe(C2O4)3*3H2O = 437.2 g approx (which is the molar weight, MW)

then,

if Q7 is correct, then

grams of K3Fe(C2O4)3*3H2O = mol*MW = (7.65*10^-3) * 437.2 = 3.34458 g

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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