When HgO is heated, it decomposes into elemental mercury and molecular oxygen gas. If 64.0 g of Hg is obtained from 85.0 g of the oxide, what is the percent yield of the reaction?

The balanced equation is 2HgO > 2Hg + 02.

2HgO--->O2+2Hg

85g(HgO)*1mol/(200.59+16)g* 2molHg/2molHgO*200.59gHg/1molHg=78.721g
that is your theoretical yield
so 64.0g/78.721*100=81.3%yeild

Well, it sounds like the HgO wanted to break free, like a rebellious teenager leaving its oxygen companion. But let's get serious for a moment and calculate the percent yield.

To find the percent yield, we need to compare the actual yield (which is 64.0 g of Hg) to the theoretical yield. Let's calculate the theoretical yield first:

Using the balanced equation, we can see that 2 moles of HgO produce 2 moles of Hg. To find the moles of HgO, we divide the given mass (85.0 g) by the molar mass of HgO, which is 216.59 g/mol.

85.0 g HgO / 216.59 g/mol = 0.3928 mol HgO

Since 2 moles of HgO produce 2 moles of Hg, the theoretical yield of Hg is also 0.3928 mol.

Now, let's convert the moles of Hg to grams:

0.3928 mol Hg * (200.59 g/mol) = 78.6 g Hg (theoretical yield)

Finally, we can calculate the percent yield:

Percent Yield = (Actual Yield / Theoretical Yield) * 100

Percent Yield = (64.0 g Hg / 78.6 g Hg) * 100

Percent Yield ≈ 81.5%

So, the percent yield of the reaction is approximately 81.5%. It seems like the reaction had a pretty good sense of humor and only slightly deviated from the expected yield.

To find the percent yield of the reaction, we need to compare the actual yield of mercury to the theoretical yield.

Step 1: Calculate the moles of HgO used.
To calculate the moles, we use the molar mass of HgO:
Molar mass of HgO = 200.59 g/mol

moles of HgO = mass / molar mass
moles of HgO = 85.0 g / 200.59 g/mol
moles of HgO = 0.4236 mol

Step 2: Determine the limiting reactant.
The limiting reactant is the reactant that gets completely consumed, limiting the amount of product formed. To determine the limiting reactant, we compare the stoichiometric ratio between HgO and Hg.

From the balanced equation, we see that 2 moles of HgO produce 2 moles of Hg.
Therefore, 0.4236 mol of HgO will produce 0.4236 mol of Hg.

Since the stoichiometry ratio is 1:1, the limiting reactant is HgO.

Step 3: Calculate the theoretical yield of Hg.
The theoretical yield is the maximum amount of product that can be obtained from the limiting reactant.

moles of Hg = moles of HgO
moles of Hg = 0.4236 mol

mass of Hg = moles of Hg x molar mass of Hg
mass of Hg = 0.4236 mol x 200.59 g/mol
mass of Hg = 85.0 g

Step 4: Calculate the percent yield.
The percent yield is the actual yield divided by the theoretical yield, multiplied by 100%.

percent yield = (actual yield / theoretical yield) x 100%
percent yield = (64.0 g / 85.0 g) x 100%
percent yield = 0.7529 x 100%
percent yield = 75.3%

Therefore, the percent yield of the reaction is 75.3%.

To find the percent yield of the reaction, you need to compare the actual yield (the amount of Hg obtained) to the theoretical yield (the amount of Hg that should have been obtained if the reaction went to completion).

First, calculate the molar mass of HgO and Hg:
- The molar mass of HgO (mercury(II) oxide) is 200.59 g/mol (200.59 g/mol = 2 * 200.59 g/mol Hg + 2 * 15.9994 g/mol O)
- The molar mass of Hg (elemental mercury) is 200.59 g/mol (since there is a 1:1 ratio between HgO and Hg)

Next, calculate the number of moles of HgO and Hg:
- Moles of HgO = mass of HgO / molar mass of HgO
- Moles of Hg = mass of Hg / molar mass of Hg

Using the balanced equation, you can determine the stoichiometry of the reaction:
- 2 moles of HgO produce 2 moles of Hg (a 1:1 molar ratio)

Calculating the theoretical yield of Hg:
- Theoretical moles of Hg = moles of HgO
- Theoretical yield of Hg = theoretical moles of Hg * molar mass of Hg

Now, calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) * 100

Plugging in the given values:
Mass of HgO = 85.0 g
Mass of Hg = 64.0 g
Molar mass of HgO = 200.59 g/mol
Molar mass of Hg = 200.59 g/mol
Theoretical moles of Hg = (85.0 g / 200.59 g/mol) = 0.4235 mol
Theoretical yield of Hg = 0.4235 mol * 200.59 g/mol = 85.0 g
Percent yield = (64.0 g / 85.0 g) * 100 = 75.3%

Therefore, the percent yield of the reaction is 75.3%.