If a solution containing 16.38 g of mercury(ii) acetate is allowed to react completely with a solution containing 5.102 g of sodium dichromate, how many grams of solid precipitate will be formed?

First, we write the balanced equation for the reaction:

Na₂Cr₂O₇ + (CH₃COO)₂Hg → 2CH₃COONa + HgCr₂O₇(s)
Next, we determine the number of moles of each substance. For this, we require the molar masses. These are:

Na₂Cr₂O₇ - 262 g/mol
(CH₃COO)₂Hg - 319 g/mol
CH₃COONa - 82 g/mol
HgCr₂O₇ - 416 g/mol

Now, the moles are:
Na₂Cr₂O₇ - 5.102 / 262 = 0.02
(CH₃COO)₂Hg - 16.38 / 319 = 0.05

From the equation, we can see that mercury (II) acetate is in excess, so the product moles will be calculated from sodium dichromate. These will be:
0.02

The mass will be:

0.02 * 416

8.32 grams of solid product

Yes she is correct. the answer is completely correct