Question

If a solution containing 25.95 g of mercury(II) chlorate is allowed to react completely with a solution containing 8.564 g of sodium dichromate, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?

Answer 1

The balance reaction of mercury(II) chlorate and sodium dichromate is as follows:

Hg(ClO₃)₂ + Na₂Cr₂O₇ ====> HgCr₂O₇(s) + 2 NaCl + 3 O₂

25.95 g of mercury(II) chlorate = 25.95 / 367.149 = 0.071 Moles

8.564 g of sodium dichromate = 8.564 / 261.96 = 0.033 Moles

Here sodium dichromate is limiting reagent.

The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

No of moles solid (HgCr₂O₇) produced = 0.033

weight of solid (HgCr₂O₇) = 0.033 x 416.578 = 13.75 gm

13.75 gm of solid precipitate (HgCr₂O₇) will be formed.

Excess will remain after the reaction =

0.071 moles – 0.033 moles = 0.038 moles in excess

0.038 moles * 367.149 g/ mole = 13.95 gm of Hg(ClO₃)₂