1. Sample Question. The elements iron and bromine react to give iron(III) bromide according to the balanced equation

2 Fe + 3 Br₂ ==> 2 FeBr₃

(a) How many moles of Br₂ are needed for complete reaction?

(b) What mass of FeBr₃ can be produced? (The molar mass of FeBr₃ is 295.56 g/mol.)

Solution. The balanced equation gives you the mole ratio for the reaction (3 moles of Br₂ to 2 moles of Fe) and for the product formed (2 moles of FeBr₃ produced for each two moles of Fe that react). By dimensional analysis:

Sc (s) + H₂SO₄ (aq)  ==>  Sc₂(SO₄)₃ (aq) + H₂ (g)

(a) What mass of Sc₂(SO₄)₃, in grams, can be obtained?

(b) What mass of hydrogen gas, in grams, can be obtained?

(c) If only 32.4 g of Sc₂(SO₄)₃ is obtained, what is the % yield of Sc₂(SO₄)₃?

Solution Hints. Hint 1: First balance the equation:

2 Sc (s) + 3 H₂SO₄ (aq)  ==> Sc₂(SO₄)₃ (aq) + 3 H₂ (g)

Hint 2: The balanced equation relates moles of reactants and moles of products. We have a mass and are asked for a mass. Thus the sequence of operations for solving the problem is

mass Sc ==> moles Sc ==> molesSc₂(SO₄)₃ ==> mass Sc₂(SO₄)₃

Solutions (by dimensional analysis)

= 38.0 g

= 0.607 g
 
 

(c) Remember that the percent yield is

= 100 x 32.4 g obtained / 38.0 g possible

= 85.3 %

Analysis of the question: we know the mass of the element and the mass of oxide that is obtained, so we can obtain the mass of the oxygen, and calculate the moles of oxygen atoms present. We know the mole ratio of oxygen to the element (5:2) from the formula given, so we will be able to calculate the moles of element. The molar mass (and thus the atomic weight) of the element can then be calculated from the mass and moles.

Mass oxygen = 3.310 g oxide - 1.854 g element = 1.456 g oxygen

Moles of oxygen atoms = (1.456 g oxygen) x (1 mol O / 16.00 g)

= 0.09100 mol O

Therefore, moles of M atoms = (0.09100 mol O) x (2 mol M / 5 mol O)

= 0.03640 mol M

The molar mass is

1.854 g M / 0.03640 mol M

= 50.93 g M /mol M

The element whose atomic weight is closest to 50.93 g/mol is vanadium, V.

2 Fe₂(SO₄)₃ (aq) + 3 Na₄P₂O₇ (aq)  ==>  Fe₄(P₂O₇)₃ (s) + 6 Na₂SO₄ (aq)

(b) What mass of Fe₄(P₂O₇)₃ can be produced?

(c) What mass of the excess reactant remains when the reaction is complete?

Formula weights (g/mol): Fe₂(SO₄)₃ 399.881; Na₄P₂O₇ 265.903; Fe₄(P₂O₇)₃ 745.213.

Solutions.

(a) We need to compare moles of Fe₂(SO₄)₃ and moles of Na₄P₂O₇.

mol Fe₂(SO₄)₃ = (4.456 g) (1 mol / 399.881 g) = 0.01114 mol

mol Na₄P₂O₇ = (3.923 g) (1 mol / 265.903 g) = 0.01475 mol

Now let's determine how much Na₄P₂O₇ is needed to react with 0.01114 mol Fe₂(SO₄)₃:

mol Na₄P₂O₇ needed:

= 0.01114 mol Fe₂(SO₄)₃ x (3 mol Na₄P₂O₇ / 2 mol Fe₂(SO₄)₃)

= 0.01671 mol Na₄P₂O₇

We don't have this much Na₄P₂O₇ so Na₄P₂O₇ is the limiting reagent.

(b) Calculate mass of Fe₄(P₂O₇)₃ that can be produced from 0.01475 mol Na₄P₂O₇:

0.01475 mol Na₄P₂O₇ x (1 mol Fe₄(P₂O₇)₃ / 3 mol Na₄P₂O₇) x 745.21 g/mol

= 3.665 g Fe₄(P₂O₇)₃

(c) Amount of Fe₂(SO₄)₃in excess = initial amount - amount reacted.

Amount reacted:

= 0.01475 mol Na₄P₂O₇ x (2 mol Fe₂(SO₄)₃/3 mol Na₄P₂O₇)

= 9.836 x 10^-3 mol Fe₂(SO₄)₃

We began with 0.01114 mol Fe₂(SO₄)₃, so the amount remaining (in grams) is

(0.01114 mol - 9.836 x 10^-3 mol) x 399.881 g Fe₂(SO₄)₃/mol

= 0.521 g
 
 

(a) and (b): See Example 4.2, pp. 155-156 in 4th edition, Kotz and Treichel

(c). How many liters of air at 25°C are required to roast 1.00 kg of silver sulfide? Assume that air contains 23% O₂ by mass and that the density of air at 25°C is 1.2 g/L. The equation for the reaction is:

Ag₂S + O₂ ==> 2 Ag + SO₂

Solution Hints for part (c). The sequence of steps to solve this problem is:

moles Ag₂S ==> moles O₂ ==> mass O₂ ==>mass of air  ==>volume of air

Note that "air contains 23% O₂ by mass" becomes

(0.23 g O₂ / g air) or, equivalently, (1 g air / 0.23 g O₂)

The density of air (1.2 g/L) means that each liter of air weighs 1.2 g, or that 1.2 g of air occupies a volume of one liter at 25°C.

Molar masses: CaSO₄ 136.14 g/mol; CaSO₄.¹/₂H₂O 145.15 g/mol

Analysis of the question. One of these compounds contains water of hydration; the other does not. Therefore we will be able to determine the amount of the hydrated compound from the mass of the water that is lost on heating.

mass H₂O lost = 5.4786 - 5.3114 g = 0.1672 g

moles CaSO₄.¹/₂ H₂O originally present

= 0.01856 mol CaSO₄.¹/₂H₂O

Now calculate the mass of CaSO₄.¹/₂H₂O and find the percentage. (Ans. 49.18%)

8. Sample Question. Lead azide, Pb(N₃)₂, is a popular detonator for commercial explosives, valued especially because of its reliability in damp conditions. It is made in a two-step reaction. First, sodium azide, NaN₃, is prepared from nitrous oxide (dinitrogen monoxide, N₂O) and metallic sodium in liquid ammonia. Then the sodium azide is allowed to react with lead nitrate in a metathesis reaction to produce lead azide. The equations are:

3 N₂O + 4 Na + NH₃ ==> NaN₃ + 3 NaOH + 2 N₂

2 NaN₃ + Pb(NO₃)₂ ==> Pb(N₃)₂ + 2 NaNO₃

Solution. There are two reactions, and the product of the first becomes a reactant in the second. Thus we apply the stoichiometric ratios in succession.

= 1102 g or 1.10 kg when properly rounded and given in the units asked for.

10. Sample Question. Both Na₂SO₄ and K₂SO₄ react with Ba(NO₃)₂ in aqueous solution to produce a precipitate of BaSO₄, which can be filtered and weighed. If 0.2784 g of a Na₂SO₄ - K₂SO₄ mixture produces 0.4018 g of BaSO₄, what is the percentage of K₂SO₄ in the mixture? Report your answer to 3 significant figures. HINT: There are two equations with two unknowns.

Molar masses (g/mol): Na₂SO₄ 142.044; K₂SO₄ 174.261; BaSO₄ 233.391

Analysis of the problem. The BaSO₄ produced tells us the total moles of Na₂SO₄ and K₂SO₄ present.

moles BaSO₄ = moles K₂SO₄ + moles Na₂SO₄ (equation 1)

mass Na₂SO₄ + mass K₂SO₄ = total mass(equation 2)

mass Na₂SO₄ = 0.2784 g - x (from equation 2)

moles K₂SO₄ = x g K₂SO₄ / (174.261 g K₂SO₄/mol)

moles Na₂SO₄ = (0.2784 - x) g Na₂SO₄ / (142.044 g Na₂SO₄/mol)

moles BaSO₄ = 0.4018 g BaSO₄ / (233.391 g BaSO₄/mol)

x = 0.1832 g K₂SO₄

% K₂SO₄ = 100 x 0.1832 g / 0.2784 g = 65.8%

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem180/CHEM_1P80_Assign_4.html
Last revised: August 20 2000 by M. F. Richardson
© Brock University, 1999