2. A student used a piece of aluminum that had a mass of 2.583 grams, and treated according to thesame procedure you used in this experiment.2 A1(s)+2 KOH(aq) + 4 H2SO4(aq) +22 H200) 2 KAI(SO4)2 12H20(s) +3 H2(g)How many moles of Al did the student use in this experiment?a.How many moles of the Alum, KAI(SO42 * 12 H20,Use the coefficients from the balanced chemical equation.)b.can be produced theoretically? (Hint-What is the mass of Alum theoretically possible? (Hint - Use the molar mass of the alumcalculated in question #1.)c.d. If the student's crystals had a mass of 32.105 g, what is the percent yield?

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Asked Sep 27, 2019
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2. A student used a piece of aluminum that had a mass of 2.583 grams, and treated according to the
same procedure you used in this experiment.
2 A1(s)+2 KOH(aq) + 4 H2SO4(aq) +22 H200) 2 KAI(SO4)2 12H20(s) +3 H2(g)
How many moles of Al did the student use in this experiment?
a.
How many moles of the Alum, KAI(SO42 * 12 H20,
Use the coefficients from the balanced chemical equation.)
b.
can be produced theoretically? (Hint-
What is the mass of Alum theoretically possible? (Hint - Use the molar mass of the alum
calculated in question #1.)
c.
d. If the student's crystals had a mass of 32.105 g, what is the percent yield?
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2. A student used a piece of aluminum that had a mass of 2.583 grams, and treated according to the same procedure you used in this experiment. 2 A1(s)+2 KOH(aq) + 4 H2SO4(aq) +22 H200) 2 KAI(SO4)2 12H20(s) +3 H2(g) How many moles of Al did the student use in this experiment? a. How many moles of the Alum, KAI(SO42 * 12 H20, Use the coefficients from the balanced chemical equation.) b. can be produced theoretically? (Hint- What is the mass of Alum theoretically possible? (Hint - Use the molar mass of the alum calculated in question #1.) c. d. If the student's crystals had a mass of 32.105 g, what is the percent yield?

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Expert Answer

Step 1

The balanced equation for the given reaction is shown below:

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2Al(s) 2KOH(aq)+4H2SO4+22H20(1)->2KAI(SO,), .12H,0(s)+ 3H2 (g)

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Step 2

Given that the mass of Aluminiumis 2.583 g. Number of moles of Aluminium can be determined by the below equation:

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No.of moles ofA=mass molarmass 2.583g 26.981g/mol 0.0957 mol

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Step 3

Ratio between the number of moles of alum and number of moles of aluminium is in the ratio 2:2 which ...

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Number of moles of alum Number of moles of aluminium Number of moles of alum=0.0957mol

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