Question

6) Liquid mercury (p = 13.56 g Hg/mL Hg) reacts with oxygen gas to form a solid mercury (II) oxide. Instructor/Tutor: a. Write the balanced chemical equation for this reaction ZHg2+O29, ~2Hg Ois) of mercury react with 5.00 g of oxygen, what is the theoretical yield, b. If 1.467 cm in grams? 5.00gu2 na 1992) = 0.1510 molaz excess in D c. What is in excess? How much excess is remaining? d. If 5.03 g of product is formed, what is the percent yield?

Answer 1

6. Reaction of liquid Mercury (Hg) with Oxygen gas (O₂) to form Mercury (II) oxide (HgO) is as shown

$Hg(l)+O_2(g)\rightarrow HgO (s)$

There are two O on left and only one on right. So to balance it out write 2 as coefficient of HgO. To further balance Hg on both sides, write 2 as coefficient of Hg.

$2Hg(l)+O_2(g)\rightarrow 2 HgO (s)$

This is the balanced chemical equation.

b. Volume of mercury given=1.467 cm³=1.467 mL

Mass of mercury=Density x volume

=13.56 g/mL x 1.467 mL=19.89 g

Molar mass of Hg=200.6 g/mol

Number of moles of Hg=Given mass/molar mass=19.89 g/200.6 g/mol=0.099 mol

Molar mass of O₂=2x16 g/mol=32 g/mol

Number of moles of O₂=Given mass/molar mass

=5.00 g/32 g/mol=0.156 mol

As per the balanced chemical equation

1 mol O₂ reacts with 2 mol Hg

So 0.156 mol O₂ will react with 2 x 0.156 mol=0.312 mol Hg

But we have only 0.099 mol Hg, so Hg is the limiting reactant and O₂ is in excess.

As per the balanced chemical equation

2 mol Hg produces 2 mol HgO (product)

So 1 mol Hg produces 1 mol HgO

And 0.099 mol Hg produces 0.099 mol HgO

Molar mass of HgO=Molar mass of Hg+Molar mass of HgO=200.6 g/mol+16 g/mol=216.6 g/mol

Mass of HgO produced=Number of moles x Molar mass

=0.099 mol x 216.6 g/mol=21.44 g

So theoretical yield=21.44 g

b) O₂ is the excess reagent as discussed above.

Also as per the balanced chemical equation

2 mol Hg reacts with 1 mol O₂

1 mol Hg reacts with 1/2 mol O₂

0.099 mol Hg reacts with (1/2)x0.099=0.0495 mol O₂

Remaining moles of O₂=0.156 mol-0.0495 mol=0.1065 mol

Mass of O₂ left=Number of moles of O₂ left x molar mass

=0.1065 mol x 32 g/mol=3.408 g≈3.41 g

d. Percent yield=(Actual yield/Theoretical yield) x100

=(5.03 g/21.44 g)x100=23.46%