1.) Silver Chromate may be prepared by the following reaction. 2 AgNO 3 (aq) + K
ID: 850322 • Letter: 1
Question
1.) Silver Chromate may be prepared by the following reaction.
2 AgNO3(aq) + K2CrO4(aq) ----> Ag2CrO4(s) + 2 KNO3(aq)
Calculate the mass of silver chromate produced when 0.623 L of 1.345 M silver nitrate is reacted with 245 mL of 1.89M potassium chromate.
2.) Aluminum Oxide is formed by reaction of excess oxygen with aluminum.
4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
Calculate the mass of aluminum required to form 18.2 kg of Aluminum Oxide i the reaction has a percent yeild of 83.4%.
3.) Aluminum oxide is an important starting material for making cryolite, Na3AlF6 (a compound used to give fireworks their yellow color) First the Aluminum oxide is synthesized from aluminum and then reacted with sodium hydroxide and hydrogen fluoride gas. If 2.6 Mg of Aluminum is reacted and all other reactants are in excess, calculate the mass of cryolite formed.
4 Al(s) + 3 O2(g) -----> 2 Al2O3(s)
Al2O3(s) + 6 NaOH(aq) +12 HF(g) ---> 2 Na3AlF6(s) + 9 H20 (l)
Explanation / Answer
1.
Consider the equation
2 AgNO3 + K2CrO4 ---> Ag2CrO4(s) + 2 KNO3(Aq) ....... (1)
2 mole of AgNO3 produces 1 mole of Ag2CrO4
Calculate moles of AgNO3 pressent available upon mixing 0.623 L of 1.345 M AgNO3 with 245 mL of 1.89 M K2CrO4
(Initial concentration of AgNO3) X (Initial volume in L) = (Final concentration of AgNO3) X (Final volume in L)
(1.345 M) X (0.623 L) = (Final concentration of AgNO3) X (0.623 L of AgNO3 + 245 mL of K2CrO4)
Converting al volumes in Liters
(1.345 M) X (0.623 L) = (Final concentration of AgNO3) X (0.623 L of AgNO3 + 0.245 L of K2CrO4)
(1.345 M) X (0.623 L) = (Final concentration of AgNO3) X (0.868 L of total solution)
(Final concentration of AgNO3) = (1.345 M) X (0.623 L) / (0.868 L of total solution)
(Final concentration of AgNO3) = 0.965 M of AgNO3
2 AgNO3 + K2CrO4 ---> Ag2CrO4(s) + 2 KNO3(Aq) --- 1
2 mole of AgNO3 produces 1 mole of Ag2CrO4
0.965 mole of AgNO3 will produce 0.965/2 mole of Ag2CrO4
0.965 mole of AgNO3 will produce 0.483 mole of Ag2CrO4
0.483 mole of Ag2CrO4 is converted to grams of Ag2CrO4 by multiplying with molecular weight of Ag2CrO4
Molecular weight of Ag2CrO4 = 331.7 g/mole
0.483 mole of Ag2CrO4 = (331.7 g/mole X 0.483 mole) of Ag2CrO4
0.483 mole of Ag2CrO4 = 160.2 g of Ag2CrO4
Mass of Ag2CrO4 produced upon mixing 0.623 L of 1.345 M AgNO3 with 245 mL of 1.89 M K2CrO4 = 160.2 g
2.
Consider the equation
4 Al (s) + 3O2 ---> 2 Al2O3 (s)
4 mole of Al produces 2 mole of Al2O3
The yield of the reaction is 84.4 %
Therefore to get 18.2 Kg of Al2O3 total produce of Al2O3 = (18.2 Kg X 100) / 84.4
18.2 Kg of Al2O3 total produce of Al2O3 = 21.6 Kg
Calculate the mole of Al2O3 in 21.6 Kg of Al2O3
Molecular weight of Al2O3 = 101.96 g/mole
mole of Al2O3 in 21.6 Kg of Al2O3 = (21.6 Kg) / 101.96 g/mole
mole of Al2O3 in 21.6 Kg of Al2O3 = (21.6 X1000 g) / 101.96 g/mole
mole of Al2O3 in 21.6 Kg of Al2O3 = 211.85 mole of Al2O3
In the equation
4 Al (s) + 3O2 ---> 2 Al2O3 (s)
2 mole of Al2O3 (s) is produced from 4 mole of Al (s)
Therefore 211.85 mole of Al2O3 will be produced from 4/2 X 211.85 mole of Al (s)
211.85 mole of Al2O3 will be produced from 423.7 mole of Al (s)
Molar mass of Al = 26.98 g/mole
Mass of Al in 423.7 mole of Al (s) = 26.98 g/mole X 423.7 mole of Al
Mass of Al in 423.7 mole of Al (s) = 11431.4 g of Al
Mass of Al in 423.7 mole of Al (s) = 11.4 Kg of Al
3.
Consider the equation
4 Al (s) + 3O2 ---> 2 Al2O3 (s)
Al2O3 (s) + 6NaOH (aq) + 12 HF (g) ---> 2 Na3AlF6 (s) + 9 H2O (l)
4 mole of Al produces 2 mole Na3AlF6
Calculate number of mole of Al in 2.6 mg of Al
Molecular weight of Al = 26.98 g/mole
Number of mole of Al in 2.6 mg of Al = 2.6 mg / (26.98 g/mole) of Al
Number of mole of Al in 2.6 mg of Al = 0.0026 g / (26.98 g/mole) of Al
Number of mole of Al in 2.6 mg of Al = 9.6 X 10-5 mole of Al
In the equation
Consider the equation
4 Al (s) + 3O2 ---> 2 Al2O3 (s)
Al2O3 (s) + 6NaOH (aq) + 12 HF (g) ---> 2 Na3AlF6 (s) + 9 H2O (l)
4 mole of Al produces 2 mole Na3AlF6
Therefore
9.6 X 10-5 mole of Al will produce 2/4 X 9.6 X 10-5 mole of Na3AlF6
9.6 X 10-5 mole of Al will produce 4.8 X 10-5 mole of Na3AlF6
Calculate the mass of 4.8 X 10-5 mole of Na3AlF6
Molar mass of Na3AlF6 = 209.94 g/mole
Mass of 4.8 X 10-5 mole of Na3AlF6 = 4.8 X 10-5 mole X 209.94 g/mole Na3AlF6
Mass of 4.8 X 10-5 mole of Na3AlF6 = 0.01 g Na3AlF6
2.6 mg of Al will produce 0.01 g Na3AlF6
2.6 mg of Al will produce 10 mg of Na3AlF6
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