Question

How many grams of aluminum sulfide would form if 2.25 liters of a 1.50 molar aluminum chloride solution reacts with 2.00 liters of a 2.75 molar hydrosulfuric acid (H2S) solution?

2AlCl3 + 3H2S Al2S3 + 6HCl (2 points)
• 18.6 g
• 101 g
• 253 g
• 275 g

Answer 1

Do you know how molar concentration of a substance in a solution is defined?

Answer 2

no.......

Answer 3

Alright. Well, it's defined as the amount ( in moles) of some substance per volume (in liters).

So if you have 1 mole of substance A in 1 liter of a solution, the molar concentration of A is...\[[A]=\frac{1mol}{1L}=1M\]"M" is shorthand for mol/L.

Make sense?

Answer 4

oh yesss

Answer 5

So, if we look back at the original problem, you have 2.25 liters of a 1.50M solution of aluminum chloride. Any idea how to convert this into moles of aluminum chloride?

Answer 6

plug it into the equation?

Answer 7

Try it out and see what you get.

Answer 8

okay hold on(:

Answer 9

A=1.50/2.25= is this right?

Answer 10

No. This is what the equation is:\[c_i=\frac{n_i}{V_{sol'n}}\]Where c_i is the molar concentration of substance i in solution, n_i is the number of moles of i in solution, and V is the volume of the solution. Does that help?

Hint: follow your units

Answer 11

im so confused.... lol

Answer 12

So from the problem, we are trying to find n, the number of moles. If we rearrange the above equation to solve for n, we get...\[c=\frac{n}{V} \rightarrow n=(c)(V)\]Does that make any sense?

Answer 13

sort of

Answer 14

Well if we plug in our numbers into that equation:\[n=(1.50M)(2.25L)=(1.50\frac{mol}{L})(2.25L)\]Liters will cancel out and we are left with:\[3.375mol\]Ok?

Answer 15

okay so then we use the other equation you posted earlier? with this new info.

Answer 16

Well, first you need to convert both of the reactants given into moles of reactants. Then, you'll want to find the limiting reactant. Do you know how to do that?

Answer 17

no this is so confusing :O

Answer 18

Alright, well let's look at the reaction equation we are given here:
\[2AlCl_3+3H_2\S \rightarrow Al_2\S_3+6HCl\]If we break this down to its most simple form, we have 2 molecules of AlCl3 reacting with 3 molecules of H2S to produce 1 molecule of Al2S2 and 6 molecules of HCl.

Does that make sense?

Answer 19

yess

Answer 20

Alright, well this also means that if we scale it up, we could have 2 moles of AlCl3 reacting with 3 moles of H2S, etc, etc.. Still keeping up?

Answer 21

yessss

Answer 22

\[(1.50M)(2.25L)=3.375mol\ AlCl_3\]\[(2.75M)(2.00L)=5.50mol\ H_2\S\]\[L.R.:AlCl_3\]\[3.375mol\ AlCl_3*\frac{1mol\ Al_2\S_3}{2mol\ AlCl_3}*\frac{150.16g\ Al_2\S_3}{1mol\ Al_2\S_3}=253.4g\ Al_2\S_3\]
The answer is C.

Answer 23

thank youuu. im writing this down in my notebook (: