Calculate the theoretical yield, percent yield, and atom economy of the synthesi

Question

Calculate the theoretical yield, percent yield, and atom economy of the synthesis reaction

2Al(s)+ 2KOH(aq) + 4H2SO4(aq) + 22H2O(l) =====> 2[KAl(SO4)2 * 12H2O](s) + 3H2(g)

Mass of Al is 1.000 g

Volume of KOH is 49.0 mL, KOH used is 1.5M

Mass of [KAl(SO4)2 * 12H2O](s) is .246 g ( This was experimentaly found and is not accurate, but disregard that solve the problem as if it was correct)

Molecular weight of [KAl(SO4)2 * 12H2O](s) is 474.37 grams/mole.

15 mL of 9M H2SO4(aq) is added

The following process is used

2Al(s)+ 2KOH(aq) + 6H2O(l) + heat =====> 2KAl(OH)4(aq)+ 3H2(g)

2KAl(OH)4(aq)+ H2SO4(aq) =====> 2Al(OH)3(s) + K2SO4(aq) + 2H2O(l)

2Al(OH)3 + 3H2SO4(aq) + heat =====> Al2(SO4)3(aq) + 6H2O(l)

Al2(SO4)3(aq) + K2SO4(aq) + 24H2O(l) =====> 2[KAl(SO4)2 * 12H2O](s)

Please posted questions if you are unsure about anything and I will ansewr them

Explanation / Answer

Chemical reaction:
2Al(s)+ 2KOH(aq) + 4H2SO4(aq) + 22H2O(l) --> 2[KAl(SO4)2 * 12H2O](s) + 3H2(g)

Mass of Al is 1.000 g (atomic mass of aluminum 26.981539 g/mol)
Moles of Al is 0.03706

Volume of KOH is 49.0 mL, KOH used is 1.5M
nKOH = 1.5 mol/L . 49.0 ml . 1 L/1000 ml
nKOH = 0.0735 mol
Moles of KOH is 0.0735
Mass of KOH is 4.1237616 g

Mass of [KAl(SO4)2 * 12H2O](s) is 0.246 g (This was experimentaly found and is not accurate, but disregard that solve the problem as if it was correct)

Molecular weight of [KAl(SO4)2 * 12H2O](s) is 474.37 grams/mole.
Moles of [KAl(SO4)2 * 12H2O] is 5.1858x10-4

15 mL of 9M H2SO4(aq) is added (molecular mass of sulfuric acid 98.079 g/mol)
nH2SO4 = 9 mol/L . 15.0 ml . 1 L/1000 ml
nH2SO4 = 0.135 mol
Moles of H2SO4 is 0.135 mol
Mass of H2SO4is 13.240665 g

Identify the limitant reactant:
Al?
If reacts all of the aluminum (0.03706 mol of Al) also reacts 0.03706 mol KOH and 0.07412 mol of H2SO4
Both the base and the acid are in excess, so the aluminum is the limitant reactant.

If the reaction is perfect (efficiency of 100%) 0.03706 mol of [KAl(SO4)2 * 12H2O] is formed.
m[KAl(SO4)2 * 12H2O] = 0.03706 mol . 474.37 g/mol
m[KAl(SO4)2 * 12H2O] = 17.58 g (theoretical yield of mass)

Instead, was obtained 0.246 g (terrible), so:
% = (0.246 g/17.58 g).100
% = 1.4
The percent yield is 1.4 %

Atom economy of the synthesis reaction:

Mass of water
mH2O = (22/2). 0.03706 mol. 18.01528 g/mol
mH2O = 7.3441 g

Mass of acid:
mH2SO4 = 0.07412 mol . 98.079 g/mol
mH2SO4 = 7.2696 g

Mass of base:
mKOH = 0.03706 mol . 56.1056 g/mol
mKOH = 2.0793 g

General equation:
atom economy = (mass of desired product/mass of all reactants).100

Theorical (efficiency of 100%)
A% = (17.58 /1 + 2.0793 + 7.2696 + 7.3441) . 100
A% = 99.36

Experimental
A% = (0.246 /1 + 2.0793 + 7.2696 + 7.3441) . 100
A% = 1.4

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