Question

If 268.5 grams of Aluminum sulfide reacts with excess water at STP, how many liters of Hydrogen sulfide gas is produced?

Answer 1

\[Al_2S_3 + 6H_2O \rightarrow 2Al(OH)_3 + 3H_2S(g)\]

Answer 2

Molar Mass of Aluminum sulfide: 268.5g
Do I use this equation? ....
\[STP=\frac{ P_1V_1 }{ n_1T_1 }=\frac{ P_2V_2}{n_2T_2 }\]

Answer 3

nope. find the moles of the reactant, then use those to find the moles produced using stoichiometry.

Once you do that use the ideal gas law, PV=nRT, at STP (standard temperature and pressure) to find the volume (i.e. L) of product.

Answer 4

Somebody already explain what to do.

Answer 5

@aaronq:

From what I understand you're saying, I do this?

Moles of Reactant:

Al2S3 = (2 x 26.98) + (3 x 32.06) = 150.14g
6H2O = (12 x 1.008) + (6 x 16) = 108.096g

\[268.5gAl_2S_3 \times \frac{ 150.14g Al_2S_3 }{ 1 mole Al_2S_3 } \times \frac{1mole6H_2O}{1moleAl_2S_3} \times \frac{108.096g 6H_2O}{1 mole 6H_2O}\]

Answer 6

Flip the second factor of what you wrote. We're trying to cancel out the grams to get the moles of Aluminum Sulfide

Answer 7

Oh, and you messed up the second half of what you wrote above. Don't do the stoichiometry on the water, you're doing it to find how many moles of hydrogen sulfide you have.

Answer 8

@iPwnBunnies: Ohhh yes. I see now. I was under the impression that you needed to take into account the information of all of the compounds on the reactants' side, but I see that this is not the case.

Thanks for pointing that out!

This should be right, then?:

\[268.5gAl_2S_3 \times \frac{1 mole Al_2S_3}{150.14g Al_2S_3} \times \frac{3 mole H_2S}{1 mole Al_2S_3} = 5.165molH_2S\]

and I go right ahead and use PV=nRT

Answer 9

Absolutely. 8) Good job.

Answer 10

Oh good. I understand it well now. Thank You so much!