If a solution containing 33.53 g of mercury(II) acetate is allowed to react completely with a solution containing 9.718 g of sodium dichromate, how many grams of solid precipitate will form

14.58 g

Explanation:

Hg(C2H3O2)2(aq) + NaCr2O7(aq) > 2NaC2H3O2(aq) + HgCr2O7(s)

Number of moles of mercury II acetate =  33.53g/318.7 g/mol = 0.11 moles

Number of moles of sodium dichromate = 9.178 g/261.97 g/mol = 0.035 moles

Since the molar ratio is 1:1,  sodium dichromate is the limiting reactant.

Molar mass of mercuric chromate = 416.58 g/mol

Mass of 416.58 g/mol produced = 0.035 moles * 416.58 g/mol = 14.58 g

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.

it is the most commonly accepted one