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

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water