Question

If a solution containing 63.00 g of mercury(II) nitrate is allowed to react completely with a solution containing 17.796 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? 

Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. 

Answer 1

The reaction of mercury (II) nitrate with sodium dichromate can be written as

Note that NaNO3 is soluble in aqueous solution but HgCr2O7 is not soluble, hence, it is precipitated out.

Mass of mercury nitrate taken = 63.00 g

Molar mass of mercury nitrate = 324.7 g/mol

Hence, number of moles of mercury nitrate taken is

Mass of sodium dichromate taken = 17.796 g

Molar mass of sodium dichromate = 261.97 g/mol

Hence, number of moles of sodium dichromate taken is

Since 1 mol of mercury nitrate reacts with 1 mol of sodium dichromate in balanced reaction, the number of moles of mercury nitrate that will react with 0.067931 mol of sodium dichromate is 0.067931 mol of mercury nitrate.

Hence, Sodium dichromate is the limiting reactant and Mercury nitrate is the excess reactant.

Each mole of sodium dichromate that reacts forms 1 mol of mercury dichromate.

Hence, number of moles of mercury dichromate that will form from 0.067931 mol of sodium dichromate is 0.067931 mol.

Molar mass of mercury dichromate =

Hence, the mass of precipitate mercury dichromate is

Hence, the mass of precipitate formed is approximately 28.30 g. (rounded to 4 significant figures).

Number of moles of excess reactant left unreacted = starting number of moles - moles reacted.

Excess reactant is mercury nitrate.

Hence,

Number of moles of excess reactant left unreacted = 0.1940 mol - 0.067931 mol = 0.126069 mol

Molar mass of mercury nitrate = 324.7 g/mol

Hence, mass of mercury nitrate left unreacted is

Hence, the mass of reactant in excess that is left unreacted is 40.93 g (rounded to 4 significant figures).

After complete precipitation, we have the following amount of each substance

(does not ionize into ions as it is the precipitate)

(completely consumed as it is the limiting reactant)

Note that only the aqueous species present in the final solution results in ion concentration.

Hence, the concentration of Hg2+ in the final solution comes from Hg(NO3)2. Each mole of Hg(NO3)2 contains 1 mol of Hg2+.

Hence,

Na+ and NO3- are the spectator ion in the reaction as they are not precipitated.

Hence, the initial moles of NO3- is equal to the final moles. Since we had taken 0.1940 mol of Hg(NO3)2 in the beginning, and each molecule of Hg(NO3)2 gives 2 molecules of NO3-, the number of moles of NO3- is

Similarly, starting concentration of Na2Cr2O7 will equal that of Na+ as Na+ is spectator ion. Each molecule contains 2 Na+ ion

Hence,

dichromate ion is completely consumed as it precipitates out.

Hence,

Hence, the answers are

Note: each number of moles is rounded to 4 significant figures. Take more significant figures in your answer as give above if necessary.