Question

QUESTION 1

Calculate the mass of SO₃ formed when 9.42 g of SO₂ reacts with 4.50 g of O₂ according to the following reaction.

                      2SO₂ + O₂ → 2SO₃

a.		22.52 g
	b.	11.77 g
c.		11.25 g
	d.	34.27 g
	e.	5.63 g

QUESTION 2

If 12.002g of potassium oxalate hydrate is added to a beaker and then 20mL of DI water is added to dissolve and then 8.0mL of 1.5 M FeCl₃ is added. Which statement is correct

	a.	the beaker contains 0.173 moles of potassium oxalate hydrate and 0.012 moles of iron(III) chloride
	b.	the beaker contains 0.0651 moles of potassium oxalate hydrate and 0.012 moles of iron(III) chloride
	c.	the beaker contains 3.255 moles of potassium oxalate hydrate and 0.012 moles of iron(III) chloride
	d.	the beaker contains 0.012 moles of potassium oxalate hydrate and 0.0651 moles of iron(III) chloride
	e.	the beaker contains 3.255 moles of potassium oxalate hydrate and 12.0 moles of iron(III) chloride

QUESTION 3

The reaction of 35.0 g of Al with excess g of Cr₂O₃ produced 25.6 g of Cr. What is the percent yield for this reaction?

                      2Al + Cr₂O₃ → Al₂O₃ + 2Cr

	a.	51.0%
	b.	196.3%
	c.	73.1%
	d.	75.9%
	e.	37.9%

QUESTION 4

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.

                      MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

            According to the above reaction.

If 1.5 moles of MnO₂ are reacted with 5.6 moles of HCl which of the following statements is true?

a.		HCl is the limiting reactant and 11.2 mol of water is produced
b.		HCl is the limiting reactant and 2.8 mol of water is produced
c.		MnO2 is the limiting reactant and 3.0 mol of water is produced
d.		HCl is the limiting reactant and 5.8 mol of water is produced
e.		MnO2 is the limiting reactant and 5.8 mol of water is produced

Answer 1

Question 1: The reaction is SO₂ + 1/2 O₂ -----> SO₃

Mass of SO2 = 9.42 g

molar mass of SO2 = 64 g/mol

moles of SO2 = mass/molar mass = 9.42/64 = 0.147 mol

Mass of O2 = 4.50 g

molar mass of O2 = 32 g/mol

moles of O2 = mass/molar mass = 4.50/32 = 0.141 mol

In reaction, for 1 mole of SO2, 0.5 mol of O2 is used.

So, for 0.147 mol of SO2, moles of O2 used = 0.5*0.147 = 0.074 mol

So, O2 is in excess in reaction and SO2 is limiting reagent. The limiting reagent will decide the amount of product formed. In reaction, 1 mol of SO2 gives 1 mol of SO3. So, 0.147 mol of SO2 will give 0.147 mol of SO3.

Molar mass of SO3 = 80 g/mol

mass of SO3 formed = moles*molar mass = 0.147*80 = 11.76 g

The option b is the closest. Correct answer is b.

Question 2: Mass of potassium oxalate hydrate = 12.002 g

molar mass of potassium oxalate hydrate = 184 g/mol

moles of potassium oxalate hydrate = mass/molar mass = 12.002/184 = 0.065 mol

Concentration of FeCl3 = 1.5 M

volume of FeCl3 = 8.0 ml = 0.008 L

Moles of FeCl3 = concentration*volume = 1.5*0.008 = 0.012 mol

Option b is correct.

Question 3: Mass of Al = 35.0 g

Molar mass of Al = 27 g/mol

moles of Al = mass/molar mass = 1.30 mol

In reaction, 2 moles of Al produces 2 moles of Cr.

So, 1.30 mol of Al will produce 1.30 mol of Cr.

Molar mass of Cr = 52 g/mol

Mass of Cr produced = moles*molar mass = 1.30*52 = 67.4 g

Cr produced (theoretical yield) = 67.4 g

Cr obtained (experimental yield) = 25.6 g

% yield = (experimental/theoretical)*100 = (25.6/67.4)*100 = 37.9 %

Option e is correct.

Question 4: In the reaction, 1 mole of MnO2 reacts with 4 moles of HCl

So, 1.5 mol of MnO2 will react with: (4/1.5) = 2.67 mol of HCl

But HCl present is 5.6 mol

So, HCl is in excess and MnO2 is limiting reagent.

If MnO2 is limiting reagent, then it will decide the moles of products formed.

1 mole of MnO2 produces 2 moles of water.

So, 1.5 moles of MnO2 will produce water = 1.5*2 = 3.0 mol

Option c is correct.