Mercury reacts with oxygen to produce mercury(II) oxide.
Hg(l) + O2(g) → HgO(s)
a. Balance the reaction equation. Include physical states.
2Hg(l) + O2(g) → 2HgO(s)
b. What is the theoretical yield (in grams) of HgO if 255 g of Hg are reacted with 255 g of O2?
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To calculate the theoretical yield of HgO, we need to determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed and thus determines the maximum amount of product that can be formed.
First, let's determine the number of moles for each reactant.
The molar mass of Hg is 200.59 g/mol, so for 255 g of Hg:
Number of Hg moles = mass of Hg / molar mass of Hg
= 255 g / 200.59 g/mol
= 1.27 mol of Hg
The molar mass of O2 is 32.00 g/mol, so for 255 g of O2:
Number of O2 moles = mass of O2 / molar mass of O2
= 255 g / 32.00 g/mol
= 7.97 mol of O2
Now, let's use the balanced equation to determine the stoichiometric ratio between Hg and HgO.
From the balanced equation:
2 Hg(l) + O2(g) → 2 HgO(s)
We can see that for every 2 moles of Hg, 2 moles of HgO are produced.
Therefore, the number of moles of HgO that can be formed will be equal to half the number of moles of Hg:
Number of moles of HgO = 1.27 mol / 2
= 0.635 mol of HgO
Finally, let's calculate the theoretical yield of HgO in grams.
The molar mass of HgO is 216.59 g/mol, so for 0.635 mol of HgO:
Mass of HgO = number of moles of HgO * molar mass of HgO
= 0.635 mol * 216.59 g/mol
= 137.53 g of HgO
Therefore, the theoretical yield of HgO if 255 g of Hg are reacted with 255 g of O2 is 137.53 grams of HgO.
To find the theoretical yield of HgO, we need to determine the limiting reactant first.
1. Calculate the number of moles for each reactant:
The molar mass of Hg is 200.59 g/mol, and the molar mass of O2 is 32.00 g/mol.
Moles of Hg = mass of Hg / molar mass of Hg = 255 g / 200.59 g/mol = 1.273 mol
Moles of O2 = mass of O2 / molar mass of O2 = 255 g / 32.00 g/mol = 7.969 mol
2. Use the balanced equation to determine the stoichiometric ratio between Hg and HgO:
According to the balanced equation: 2 moles of Hg = 2 moles of HgO
So, the ratio of moles of Hg to HgO is 2:2 or 1:1.
3. Determine the limiting reactant:
The reactant that is completely consumed or used up first is the limiting reactant.
Since the ratio between Hg to HgO is 1:1, we can conclude that Hg is the limiting reactant because we have 1.273 mol of Hg, which is less than 2 moles required for complete reaction.
4. Calculate the number of moles of HgO formed based on the limiting reactant:
Since 1 mol of Hg reacts to produce 1 mol of HgO, the number of moles of HgO formed is also 1.273 mol.
5. Calculate the mass of HgO formed:
The molar mass of HgO is 216.59 g/mol.
Mass of HgO = moles of HgO x molar mass of HgO
= 1.273 mol x 216.59 g/mol
= 275.83 g
Therefore, the theoretical yield of HgO is 275.83 grams when 255 grams of Hg react with 255 grams of O2.