Question

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

Answer 1

Answer:

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

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

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 :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ 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 :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ 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.