In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole product. In a second experiment, the same sulfide was produced in the reaction of 1.52g mercury and 1.02g sulfur.

Question

a) What mass of the sulfide of mercury was produced in the second experiment?
b)Which element (mercury or sulfur) remained unreacted in the second experiment?
c)What mass remained unreacted in the second experiment?

Answer 1

Balance the reaction.

Use the mole relationships is the standard way.

Now, you can use ratios. In the first experiment 1 g Hg required .16 g S, so

1/1.52=.16/S and solve for the amount of S needed, or

S=1.52(.16)=about .24 g S

so the mass made will be about 1.76 g (check that).

Answer 2

b) Sulfur

Answer 3

c) add the total amount of reactants in the 2nd rxn and subtract it by the answer you got for part a