Question

If a solution containing 59.916 g of mercury(II) perchlorate is allowed to react completely with a solution containing 16.642 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

Hg(ClO4)2 + Na2Cr2O7 --------------------> HgCr2O7 + 2 NaClO4

399.5 g 262 g 416.6 g 2*122.44 g

the above given masses are molar masses of the reactants and products

the amount of sodium dichromate required to react with 59.916g of

mercury(II) perchlorate  = 59.916 x 262 / 399.5 = 39.294 g

but here we have 16.642 g of sodium dichromate

. so Na2Cr2O7 is limiting reagentand mercury(II) perchlorate is in excess

amount of HgCr2O7 produced = 16.642 g Na2Cr2O7 * 416.6 g HgCr2O7 /  262 g Na2Cr2O7 = 26.462 g =mass of solid precipitate

the amount of sodium dichromate required to react with

the amount of mercury(II) perchlorate required to react with 16.642 g of

sodium dichromate = 16.642 * 399.5 / 262 = 25.376

mass of HgCr2O7 produced = 26.462 g = mass of solid precipitate Answer

excess Hg(ClO4)2 remains = 59.916 - 25.376 = 34.540 g

excess Hg(ClO4)2 remains = 34.540 g Answer