Asked by Judy
If 48.1 g of solid Al2S3 and 6.68 mL of liquid H2O are reacted stoichiometrically according to the balanced equation, how many grams of solid Al2S3 remain?
Al2S3(s) + 6H2O(l) ¡æ 2Al(OH)3(s) + 3H2S(g)
Molar Mass (g/mol)
Al2S3 150.16
H2O 18.015
H2O 0.9982
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
Al2S3(s) + 6H2O(l) ¡æ 2Al(OH)3(s) + 3H2S(g)
Molar Mass (g/mol)
Al2S3 150.16
H2O 18.015
H2O 0.9982
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
Answers
Answered by
DrBob222
Judy---All of these questions look alike except the numbers have been changed. You show no work on your own. If you wish to post ONE question at the top of the page and tell us how you think the problem should be solved, along with what you don't understand about the problem, perhaps we can help get you started.
Answered by
Judy
I don't even know how to start :/
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