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.

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?

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

To find the answers to these questions, we need to use the concept of stoichiometry and the law of conservation of mass.

First, let's determine the chemical equation for the reaction between mercury and sulfur to form the sulfide of mercury. The balanced equation for this reaction is:

Hg + S → HgS

a) To find the mass of the sulfide of mercury produced in the second experiment, you need to compare the amounts of mercury and sulfur used in the reaction with the ratio provided by the balanced equation.

In the second experiment, we have 1.52 g of mercury and 1.02 g of sulfur. We can convert the masses of mercury and sulfur into moles using their molar masses.

Molar mass of Hg = 200.59 g/mol
Molar mass of S = 32.06 g/mol

Number of moles of Hg = 1.52 g Hg / 200.59 g/mol = 0.00758 mol Hg
Number of moles of S = 1.02 g S / 32.06 g/mol = 0.0318 mol S

Now, we compare the mole ratio between Hg and HgS from the balanced equation. It shows that 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS. Therefore, the number of moles of HgS produced should be equal to the number of moles of S used.

Number of moles of HgS produced = 0.0318 mol HgS

Finally, we can convert the number of moles of HgS into grams using its molar mass.

Molar mass of HgS = 232.6 g/mol

Mass of HgS produced = 0.0318 mol HgS x 232.6 g/mol = 7.39 g

Therefore, in the second experiment, a) 7.39 g of the sulfide of mercury was produced.

b) To determine which element (mercury or sulfur) remained unreacted in the second experiment, we can compare the amount used with the amount required according to the balanced equation.

In the second experiment, we used 1.52 g of mercury and 1.02 g of sulfur. To find the element that remained unreacted, we need to compare these amounts with their stoichiometric ratios.

From the balanced equation, we have a 1:1 mole ratio between Hg and S. This means that for every mole of mercury, we need one mole of sulfur. Therefore, if the stoichiometry is not satisfied, the element with excess will be unreacted.

In this case, since we have an excess of sulfur (1.02 g sulfur), it means that the sulfur remained unreacted in the second experiment.

c) To find the mass of the unreacted element, sulfur in this case, we simply subtract the mass of the sulfur involved in the reaction from the total mass of sulfur used.

Mass of sulfur unreacted = Mass of sulfur used - Mass of sulfur involved in the reaction
Mass of sulfur unreacted = 1.02 g - 1.02 g = 0 g

Therefore, c) all of the sulfur reacted in the second experiment, and there was no sulfur remaining unreacted.

b) Sulfur

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).