Calculate the maximum number of moles and grams of H2S thatcan form when 181g al

Question

Calculate the maximum number of moles and grams of H2S thatcan form when 181g aluminum sulfide reacts with 118g water. Al2S3+6H2O --> 2Al(OH)3 + 3H2S I figured out the mol (3.27) and grams (111.61) of H2S but Ican't seem to get this question right What mass of the excess reactant remains? Calculate the maximum number of moles and grams of H2S thatcan form when 181g aluminum sulfide reacts with 118g water. Al2S3+6H2O --> 2Al(OH)3 + 3H2S I figured out the mol (3.27) and grams (111.61) of H2S but Ican't seem to get this question right What mass of the excess reactant remains?

Explanation / Answer

Hmmm, did you find the right limiting reactant? To solve this question, note that the reaction requires 6 times theH2O compared to your Al2S3. It looks like you understand how to solve for how muchH2S is produced, so I'll make this short. Find out how many moles of Water and Aluminum Sulfide are used,then use that information to find out which produces less HydrogenSulfide. Remember, the Aluminum Sulfide is every 1 mole of Aluminumproduces 3 of Hydrogen Sulfide, whereas Water produces 3 HydrogenSulfide for every 6 Water. Feel free to message me if you still feel confused.

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