How many grams of AlCl3 (Molar mass =133.5 g/mol) are needed to prepare 125 mL of a 0.150 M solution?

A) 20.0 g AlCl3

B) 16.7g AlCl3

C) 2.50 g AlCl3

2.50 g of AlCl3

Explanation:

Goodness, stoichiometry...

So, what we need to find first is the amount of grams of AlCl3. To do this we look at the formula of molarity.

M = mols/L of solvent

So we know two parts of this formula. We have the Molarity (0.150) and the mL (125).

Now, we can't forget that we must convert 125 mL into liters so we have 0.125 L ( I forgot and had to do the entire problem again...)

So if we do the backwards equation we get:

0.150 = x/0.125

If we do math (fun ikr) we get 18.75 mols of the solution.

Now, we have to plug this wonderful number into stoichiometry

0.01875 mols | 133.5 g

                  | 1 mol AgCl3

If you are unfamiliar with what I'm doing, I'm basically going to multiply 0.01875*133.5 then divide that whole thing by 1.

So, I got 2.503125 g AlCl3

If your teacher is a stickler for significant figures there are 3 sig figs for this problem so your final answer should be

2.50 g of AlCl3

Hope you have a great day and fun with chemistry

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation:

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g