Vermilion, a very rare and expensive solid natural pigment that has been used to print artists' signatures on works of art, is mercury(ii) sulfide. what mass of vermilion can be produced when 75.0 ml of 0.02700 m mercury(ii) nitrate is mixed with a solution containing excess sodium sulfide? vermilion is insoluble in water.

To solve for moles use molar and volume:

M = mol/L

0.02700 m= mol/(.75 L)

mol = .036 mol of Hg(NO3)2

The proportion of Hg(NO3)2 to the Mercury that is used to create HgS is 1:1 so solve for the moles of Mercury used. So:

.036 * (1 Hg(NO3)2 / 1 Hg) = .036 moles of Hg used to make HgS

The proportion of Mercury to HgS is 1:1, as an alternative of doing the obvious math you can conclude that .036 moles of HgS will be produced because you're given .036 moles of Hg and an excess of S. Use the molar mass of HgS to determine how many grams will be produced.

.036 moles * 232.66 g/mol = 8.37576 grams of Vermilion is produced.

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