If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of sodium dichromate, how many grams of solid precipitate will form

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

Moles of mercury(II) acetate =

Moles of sodium dichromate =

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.