A sample of 0.72g of Al is reacted with 17mL 2.4M KOH. Which one is the limiting reagent?

2Al+2KOH+6H2O------------->2KAl(OH)4+3H2

I don't know how to find limiting reagent on this case.. I think I need to know sample gram of KOH... please help me how to find out sample gram of KOH..

Thank you so much..

If you teach me how to find sample gram of KOH, then I can find limiting reagent,

If your equation is balanced (it looks like it is), you need 2 moles of Al per equivalent; 2 moles of KOH per equivalent, and 6 moles of H2O per equivalent. I guess water is in excess since none is given. Calculate the moles of Al and the moles of KOH. Convert to equivalents.

This one is particularly easy since the number of moles of Al is the same as the number of moles of KOH that must react.

Al=(.72g)(1mol/26.98g)=0.0267moles

How can I find mole of KOH?

You have the concentration and volume :rolleyes:

So.. KOH mole is .408mole?

:eek:

Hmm, there must have been a mistake in your calculation, I got 0.0408 mol.

ohhh yes you are right I put 24mole instead of 2.4M

I got 0.0408moles of KOH

Finally I got 0.01335 equivalents of Al and 0.0204 equivalents of KOH.

So, it means Al is limithing reagent...

I hope my calculation are correct..

^^*

;)

Al truly is the limiting reagent. However, I don't understand why you looked for 0.01335 and 0.0204 moles.

1 mole of Al reacts with 1 mole of KOH
You have 0.0267 moles of Al, which would react with 0.0267 moles of KOH
But you have 0.0408, which is a lot more than 0.0267 that you require. So, from there, you say that Al is limiting.

Ohh that way is much easy!

Thanks

If a salt has solubility of 27.2g/100g water at 25ºC, what voulme of water would be needed to dissolve 12.4g of the salt.

Do you know any volume formula for find this answer?

Also do you know theoretical yield formula?
Calculate the theoretical yield of KAl(OH)4 formed in very top question(main question) if 1g of Al reacts completely with KOH.

Please help me..

Well, solubility:

1. You know that 27.2 g of salt dissolves completely in 100 g of water.
So, 1 g would completely dissolve in 100/27.2 g of water
12.4g would completely dissolve in 100/27.2 * 12.4 = 45.6g of water.

Now 1 g of water has a volume of 1 cm^3 or 1 mL
You need 45.6 g of water, which is equivalent to 45.6 mL of water.

All this, assuming that the temperature is 25 C.

2. I know it as yield, and I don't use any formula either.
You know that you have 0.0267 mol of KOH reacting (the other are in excess). Also, your equation tells you that 2 molea of KOH react to give 2 moles of KAl(OH)4, that is the mole ratio is 1:1.

So, you will have 0.0267 mol of KAl(OH)4 being formed.
What is the mass of this KAl(OH)4? That mass is the yield.

Now, if there is percentage yield, you divide the actual weight of the product by the theorical yield, multiplied by 100%.

3.58g/1g x 100% = 358%?

No, the opposite. You got 1 g, that's the actual weight. Theorically, it should be 3.58 g. So, it becomes 1/3.58 x 100% = 27.9%