A student reacts 4.991 g of Fe(NH 4 ) 2 (SO 4 ) 2 * 6 H 2 O with 26.2 mL of 1 M

Question

A student reacts 4.991 g of Fe(NH4)2(SO4)2 * 6 H2O with 26.2 mL of 1 M H2C2O4 solution to produce the yellow solid, FeC2O4 * 2 H2O. This solid is then reacted with 15 mL of saturated K2C2O4 * H2O solution (300 g K2C2O4 * H2O per liter of solution), 20 mL 3% H2O2, and 8 mL of 1.0 M H2C2O4.

1. Determine the moles of Fe(NH4)2(SO4)2 * 6 H2O added.

2. Determine the moles of H2C2O4 added (initially).

3. Determine the moles of K2C2O4 * H2O added. (Note --- You will need to use the fact that there are 300 g K2C2O4 * H2O per liter of solution.)

4. Based on your answers from parts 1-3, what is the limiting reagent? (Assume all other reagents are in excess.)

5. Based on the limiting reagent above, determine the theoretical yield?

6. If the student obtains 4.881 g of product (K3Fe(C2O4)3 * 3 H2O), what is the percent yield for the reaction?

Please show all steps in the calculation Thanks

Explanation / Answer

A student reacts 4.991 g of Fe(NH 4 ) 2 (SO 4 ) 2 * 6 H 2 O with 26.2 mL of 1 M

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