I have 12g of Aluminum. 420 mL of 2.0 M KOH. 216 mL of 10.0 M H2SO4. 300mL oh 50:50 ethanol. The original balanced equation shows 2 moles of Aluminum, 2 moles KOH, 6 moles H2O and later on states 4 moles H2SO4. I am supposed to be creating alum [KAl(SO4)2*12H2O]. However, with the current amounts of reactants, it will make much more alum than I need. I need only 23g of Alum and the materials I have are of different molarities such as 95% ethanol, 1.4 M KOH and 9.0 M H2SO4. What I am basically asking is how to I scale down these amounts in order to produce only 23g of alum? The ending balanced equation of the UNSCALED reactants and products is such:
2Al + 2KOH + 4H2SO4 + 22H2O --> 2KAl(SO4)2*12H2O + 3H2
The first question I am asked is
1. what is the limiting reagent.
2. how much aluminum is necessary to produce 23g of alum.
3. By what factor must the mole amount of Al in the given procedure be reduced in order to produce 23g of alum?
4. What is the result of scaling the number of moles of KOH used in the given procedure by the factor determined in question 4? What volume of 1.4M KOH contains this number of moles?
(so on and so forth for the rest of the reactants).
I would like to know if there is a general equation to know for this, I am looking to understand how this is performed, as I have all the answers, but I have no idea what steps to go through to get them.
Thank you, I really hope someone can help me figure this out..
P.S., this is Lab 3 of CHEM 2070 from Cornell University, if anyone was wondering.
2Al + 2KOH + 4H2SO4 + 22H2O --> 2KAl(SO4)2*12H2O + 3H2
The first question I am asked is
1. what is the limiting reagent.
2. how much aluminum is necessary to produce 23g of alum.
3. By what factor must the mole amount of Al in the given procedure be reduced in order to produce 23g of alum?
4. What is the result of scaling the number of moles of KOH used in the given procedure by the factor determined in question 4? What volume of 1.4M KOH contains this number of moles?
(so on and so forth for the rest of the reactants).
I would like to know if there is a general equation to know for this, I am looking to understand how this is performed, as I have all the answers, but I have no idea what steps to go through to get them.
Thank you, I really hope someone can help me figure this out..
P.S., this is Lab 3 of CHEM 2070 from Cornell University, if anyone was wondering.
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You are given 0.445 mol Al (12 g/26.98 g/mol)
You are given 0.840 mol of KOH (0.420 L x 2.0 mol/L = 0.840 mol)
You are given 2.160 mol of H2SO4 (0.216 L x 10.0 mol/L = 2.16 mol)
The ratio of Al : KOH : H2SO4 in the balanced equation is 2 : 2 : 4 (1 : 1 : 2)
(1)
Al is the limiting reagent
(2)
The molar mass of KAl(SO4)2.12H2O is 474.4
23 g would be 0.04848 mol
You need 0.04848 mol of Al (moles of Al = moles of KOH = moles of Alum)
0.04848 mol x 26.98 g/mol = 1.308 g of Al (to produce 0.0484 mol of Al)
You should be able to do the rest...you need 0.04848 mol Al, 0.04848 mole of KOH and 0.09696 mole of H2SO4 (maintain the ratio of 1 : 1 : 2)
You are given 0.840 mol of KOH (0.420 L x 2.0 mol/L = 0.840 mol)
You are given 2.160 mol of H2SO4 (0.216 L x 10.0 mol/L = 2.16 mol)
The ratio of Al : KOH : H2SO4 in the balanced equation is 2 : 2 : 4 (1 : 1 : 2)
(1)
Al is the limiting reagent
(2)
The molar mass of KAl(SO4)2.12H2O is 474.4
23 g would be 0.04848 mol
You need 0.04848 mol of Al (moles of Al = moles of KOH = moles of Alum)
0.04848 mol x 26.98 g/mol = 1.308 g of Al (to produce 0.0484 mol of Al)
You should be able to do the rest...you need 0.04848 mol Al, 0.04848 mole of KOH and 0.09696 mole of H2SO4 (maintain the ratio of 1 : 1 : 2)