Chemistry Problem

Question

f a solution containing 39.689 g of mercury(II) chlorate is allowed to react completely with a solution containing 12.026 g of sodium dichromate, how many grams of solid precipitate will form?

How many grams of the reactant in excess will remain after the reaction?

Answer 1

Mercury (II) chlorate = Hg(ClO₃)₂

Sodium dichromate = Na₂Cr₂O₇

Hg(ClO₃)₂ + Na₂Cr₂O₇ ==> HgCr2O7(s) + 2NaClO3(aq)

Find the limiting reactant: One way is to divide mols of each reactant by the corresponding coefficient in the balanced equation and which ever value is less represents the limiting reactant

For Hg(ClO3)2: 39.689 g x 1 mol / 367.50 g = 0.099834 moles (÷1->0.09)

For Na2Cr2O7: 12.026 g x 1 mol / 261.97 g = 0.04591 moles (÷1->0.046)

Since 0.046 is less than 0.09, the Na2Cr2O7 is limiting and will determine the amount of HgCr2O7(s)

Grams HgCr₂O₇(s) = 0.04591 mol Na₂Cr₂O₇ x 1 mol HgCr₂O₇(s)/mol Na₂Cr₂O₇ x 416.58 g/mol = 19.124 g

Moles Hg(ClO3)2 used up = 0.04591 mol since reacts 1:1 with Na2Cr2O7

Moles Hg(ClO3)2 left over = 0.099834 - 0.04591 = 0.05392 moles

Grams left over = 0.05392 mols x 367.50 g / mol = 19.817 g left over