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 1.50 g mercury and 1.08 g sulfur . What mass of the sulfide of mercury was produced in the second experiment?
find the formula of the sulfide
You know 1.00 g of the compound is Hg.
Hg = 1/molmassHg = number moles H=
= 1/200 moles= 0.005 moles Hg
S= .16/32 moles= .005 moles S
To the mole ratio of Hg to S is 1:1, so the empirical formula is HgS
Now, in the second, you have 1.08g S, figure the number of moles of that. Then, you have the same number of moles of Hg. What is the mass of that number of moles of Hg? The mass of the sulfide then is the Hg mass, plus 1.08g
To find the mass of the sulfide of mercury produced in the second experiment, we need to set up a proportion comparing the masses of mercury and sulfur in both experiments.
Let's call the mass of the sulfide of mercury "x" grams.
In the first experiment:
Mass of mercury = 1.00 g
Mass of sulfur = excess (unknown)
In the second experiment:
Mass of mercury = 1.50 g
Mass of sulfur = 1.08 g
Since the reaction produces the same sulfide of mercury in both experiments, we can set up the following proportion:
1.00 g mercury / x g sulfide = 1.50 g mercury / 1.08 g sulfur
To solve for "x," we can cross-multiply and solve the equation:
1.00 g mercury * 1.08 g sulfur = 1.50 g mercury * x g sulfide
1.08 g sulfur = 1.50x g sulfide
Dividing both sides by 1.50:
1.08 g sulfur / 1.50 = x g sulfide
x = 0.72 g
Therefore, the mass of the sulfide of mercury produced in the second experiment is 0.72 grams.
To find the mass of the sulfide of mercury produced in the second experiment, let's break down the problem into steps:
Step 1: Calculate the molar mass of the sulfide of mercury.
The molar mass of a compound is the sum of the molar masses of each element present. The molar mass of mercury (Hg) is 200.59 g/mol, while the molar mass of sulfur (S) is 32.06 g/mol. Since the sulfide of mercury contains one mercury atom and one sulfur atom, its molar mass is 200.59 g/mol + 32.06 g/mol = 232.65 g/mol.
Step 2: Calculate the number of moles of mercury and sulfur used in the second experiment.
Using the given mass of mercury (1.50 g) and its molar mass (200.59 g/mol), we can calculate the number of moles of mercury:
1.50 g / 200.59 g/mol = 0.00749 mol
Similarly, using the mass of sulfur (1.08 g) and its molar mass (32.06 g/mol), we can calculate the number of moles of sulfur:
1.08 g / 32.06 g/mol = 0.0337 mol
Step 3: Determine the limiting reactant.
To determine the limiting reactant, compare the ratios of the moles of mercury and sulfur used in the second experiment to the stoichiometric ratio of the reaction. The balanced equation for the reaction between mercury and sulfur is:
Hg + S → HgS
From the balanced equation, we can see that one mole of mercury reacts with one mole of sulfur to produce one mole of sulfide of mercury (HgS).
Based on the stoichiometry, the ratio of moles of mercury to sulfur is 1:1. In the second experiment, we have 0.00749 mol of mercury and 0.0337 mol of sulfur.
Since the stoichiometric ratio of moles of mercury to sulfur is 1:1, the limiting reactant is the one that is present in a lower mole ratio. In this case, mercury is the limiting reactant because it is present in a smaller mole ratio.
Step 4: Calculate the actual mass of the sulfide of mercury produced.
Now that we know that mercury is the limiting reactant, we can determine the number of moles of sulfide of mercury produced based on the moles of mercury used. From the stoichiometry, we know that one mole of mercury reacts to produce one mole of sulfide of mercury (HgS).
Therefore, the number of moles of sulfide of mercury produced is equal to the number of moles of mercury used, which is 0.00749 mol.
Finally, we can calculate the mass of the sulfide of mercury produced using its molar mass:
Mass of sulfide = Number of moles × Molar mass = 0.00749 mol × 232.65 g/mol = 1.74 g
Therefore, the mass of the sulfide of mercury produced in the second experiment is 1.74 g.