Question

The following equation for the reaction of alum with barium chloride is not balanced. KAl(SO4)2•12H2O(aq) + BaCl2(s) → KCl(aq) + AlCl3(aq) + BaSO4(s) a) Balance the equation. Calculate the mass (g) of barium chloride needed to react with 25 mL of a 0.10 M alum solution. b) In an experiment 1.02 grams of BaSO4(s) was produced. Calculate the percent yield of BaSO4(s).

Answer 1

Answer:

See explanation below

Explanation:

First, let's write again the reaction to balance it:

KAl(SO₄)₂.12H₂O + BaCl₂ ---------> KCl + AlCl₃ + BaSO₄ + H₂O

The principle to balance a reaction, is:

Metals > Non metals > hydrogen > oxygen

Following this priority order, let's balance the equation beggining with K, Al and Ba:

In the case of K and Al, they are both balanced, so we don't need to do nothing with them. In the case of Ba happens the same thing for now. LEt's see now the chlorine. In the reactants we have 2, and in products we have 3, so we'll balance the reactants putting a 2 in there:

KAl(SO₄)₂.12H₂O + 2BaCl₂ ---------> KCl + AlCl₃ + BaSO₄ + H₂O

This will unbalance the Ba, so, we just have to put a 2 on the other side:

KAl(SO₄)₂.12H₂O + 2BaCl₂ ---------> KCl + AlCl₃ + 2BaSO₄ + H₂O

Now, looking at the sulfur, is balanced both sides (thanks to the 2 we put on the product to balance the Ba).

Finally hydrogen and oxygen, in the reactants we have 12 molecules of water and in products we have two, therefore:

KAl(SO₄)₂.12H₂O + 2BaCl₂ ---------> KCl + AlCl₃ + 2BaSO₄ + 12H₂O

Now we have the whole equation balanced.

Now, let's calculate the needed mass to react with the alum solution. With the volume and concentration of the alum, we can calculate the moles:

moles = 0.1 * 0.025 = 0.0025 moles

According to the balanced reaction, we have a mole ratio of 1:2, so this means that 1 mole of the alum requires 2 moles of BaCl₂, therefore:

mole alum / moles BaCl₂ = 1/2

moles BaCl₂ = 0.0025 * 2

moles BaCl₂ = 0.005 moles

Now to get the mass we need the molar mass, which in this case is 208.23 g/mol, so:

m = 0.005 * 208.23

m = 1.0412 g

This is the mass required to react with the alum

b) for this part, we just need to calculate how many grams of BaSO₄ is produced given the innitial volume and moles calculated in part a).

In this case, we have the same mole ratio between the BaCl₂ and BaSO₄, therefore, we will have the same moles:

moles BaSO₄ = 0.005 moles

The mass of BaSO₄ using it's molar mass reported: (233.38 g/mol)

m = 0.005 * 233.38 = 1.1669 g

Then, the %yield would be:

% = 1.02 / 1.1669 * 100

% = 87.41%

Answer 2

Answer:

A. 1.04g of BaCl2.

B. Percentage yield of BaSO4 is 87.6%

Explanation:

A. The balanced equation for the reaction. This given below:

KAl(SO4)2•12H2O(aq) + 2BaCl2(s) → KCl(aq) + AlCl3(aq) + 2BaSO4(s) + 12H2O(l)

Next, we shall determine the number of mole in 25 mL of a 0.10 M alum. This is illustrated below:

Volume = 25mL = 25/1000 = 0.025L

Molarity = 0.1M

Mole =..?

Mole = Molarity x Volume

Mole of alum = 0.1 x 0.025 = 2.5x10¯³ mol.

Next, we shall convert 2.5x10¯³ mol of alum to grams.

Number of mole alum, KAl(SO4)2•12H2O = 2.5x10¯³ mol

Molar Mass of alum, KAl(SO4)2•12H2O = 39 + 27 + 2[32+(16x4)] + 12[(2x1) + 16]

= 39 + 27 + 2[32 + 64] + 12[2 + 16]

= 39 + 27 + 2[96] + 12[18]

= 474g/mol

Mass of alum, KAl(SO4)2•12H2O =..?

Mass = mole x molar mass

Mass of alum, KAl(SO4)2•12H2O = 2.5x10¯³ x 474 = 1.185g

Next, we shall determine the mass of alum and BaCl2 that reacted and the mass of BaSO4 produced from the balanced equation. This is illustrated below:

Molar mass of alum, KAl(SO4)2•12H2O = 474g

Mass of alum, KAl(SO4)2•12H2O from the balanced equation = 1 x 474 = 474g

Molar mass of BaCl2 = 137 + (35.5x2) = 208g/mol

Mass of BaCl2 from the balanced equation = 2 x 208 = 416g

Molar mass of BaSO4 = 137 + 32 + (16x4) = 233g/mol

Mass of BaSO4 from the balanced equation = 2 x 233 = 466g

Summary:

From the balanced equation above,

474g of alum reacted with 416g of BaCl2 to produce 466g of BaSO4.

Finally, we can calculate the mass of BaCl2 needed for the reaction as follow:

From the balanced equation above,

474g of alum reacted with 416g of BaCl2.

Therefore, 1.185g of alum will react with = (1.185 x 416)/474 = 1.04g of BaCl2.

Therefore, 1.04g of BaCl2 is needed for the reaction.

B. Determination of the percentage yield of BaSO4(s).

We'll begin by calculating the theoretical yield of BaSO4. This is illustrated below:

From the balanced equation above,

474g of alum reacted to produce 466g of BaSO4.

Therefore, 1.185g of alum will react to produce = (1.185 x 466)/474 = 1.165g of BaSO4.

Therefore, the theoretical yield of BaSO4 is 1.165g.

Finally, we shall determine the percentage of BaSO4 as follow:

Actual yield of BaSO4 = 1.02g.

Theoretical yield of BaSO4 = 1.165g.

Percentage yield of BaSO4 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield of BaSO4 = 1.02/1.165 x 100

Percentage yield of BaSO4 = 87.6%