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%
