Calculating Molarity of a Solution: Aluminum Salt and Unreacted Sulphuric Acid

[zpmodel]

I have problem and can not solve it:

A piece of metallic aluminum weighing 2.50 grams is treated with 75.0 mL of sulphuric acid ( d = 1.18 g/cm3 , 24.7% H2SO4 ). After the metal is dissolved, the solution is diluted to 400mL.

a) Calculate the morality of the solution with respect to the aluminum salt formed.
b) Calculate the morality of the resulting solution with respect to unreacted sulphuric acid.


I made a slight attempt, but got lost after it:

75ml x 1.18g = 88.5g

Xg / 88.5g = .247
Xg = 21.859g of H2SO4

I actually don't even know what I am doing anymore :(
I'm seeking a learning experience so that I can solve other problems as well. Please, any help would be greatly appreciated.

 

[rockonthelawn]

I have problem and can not solve it:

A piece of metallic aluminum weighing 2.50 grams is treated with 75.0 mL of sulphuric acid ( d = 1.18 g/cm3 , 24.7% H2SO4 ). After the metal is dissolved, the solution is diluted to 400mL.

a) Calculate the morality of the solution with respect to the aluminum salt formed.
b) Calculate the morality of the resulting solution with respect to unreacted sulphuric acid.


I made a slight attempt, but got lost after it:

75ml x 1.18g = 88.5g

Xg / 88.5g = .247
Xg = 21.859g of H2SO4

I actually don't even know what I am doing anymore :(
I'm seeking a learning experience so that I can solve other problems as well. Please, any help would be greatly appreciated.

Im not sure what the morality of the solution would be...but i may be able to help with the molarity...

You have 75mL of 24.7% H2SO4. 75 X 0.247 = 18.53mL of pure H2SO4. 18.53mL X 1.18g/cm3 (density) gives 21.86g H2SO4 X 1mol / 98.078g (molecular weight of H2SO4) = 0.2229mol of H2SO4

Now we have to determine the #moles of aluminum. Assuming Aluminum is purely elemental aluminum, the molecular weight x weight given is 1 mol / 26.98g X 2.50g Al = 0.09266 mol AL

First thing to do in any problem like this is to convert everything to # of moles.

We have 0.2229mol H2SO4 and 0.09266mol Al

The rest is super easy

for part A. you know the number of moles, and amount of solution (0.400L) so 0.09266mol AL / 0.400L = 0.232M

for part B. The amount of H2SO4 remaining (assuming all Al reacted) is 0.2229-0.09266 = 0.13024mol of unreacted H2SO4. so 0.13024/ 0.400L = 0.326M H2SO4.

Hope that helps (and is correct)

 

[Blueshift5]

First, let's break down the problem and identify the key information given. We are given a piece of aluminum weighing 2.50 grams and it is treated with 75.0 mL of sulphuric acid with a density of 1.18 g/cm3 and a concentration of 24.7% H2SO4. After the reaction, the solution is diluted to 400mL.

a) The first step is to calculate the moles of aluminum used in the reaction. We can do this by using the molar mass of aluminum, which is 26.98 g/mol. So, 2.50 grams of aluminum is equal to 2.50/26.98 = 0.0927 moles of aluminum.

Next, we need to determine the volume of the final solution. We started with 75.0 mL of sulphuric acid and then diluted it to 400mL, so the final volume is 75.0 mL + 400mL = 475mL.

Now, we can use the formula for molarity, which is moles of solute divided by liters of solution. In this case, our solute is the aluminum salt and our solution is the diluted solution of 475mL. So, the molarity of the solution with respect to the aluminum salt is 0.0927 moles / 0.475 L = 0.195 M.

b) To calculate the molarity of the resulting solution with respect to unreacted sulphuric acid, we need to first determine the moles of sulphuric acid used in the reaction. We can do this by using the molarity formula again, but this time we know the moles and the volume, so we can rearrange the formula to solve for moles. 0.247 M = moles / 0.075 L, so moles of sulphuric acid used is 0.247 x 0.075 = 0.0185 moles.

Next, we need to take into account the dilution factor of the solution. We started with 75.0 mL of sulphuric acid and then diluted it to 400mL, so the dilution factor is 400/75 = 5.33. This means that the moles of sulphuric acid used in the reaction is actually 5.33 times more than the calculated value of 0.