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

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

no.......

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?

oh yesss

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?

plug it into the equation?

Try it out and see what you get.

okay hold on(:

A=1.50/2.25= is this right?

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

im so confused.... lol

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?

sort of

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?

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

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?

no this is so confusing :O

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?

yess

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?

yessss

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

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

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