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A mixture of 15.0 g of aluminum sulfide and 10.0 g of water reacts according to the following equation:

Al2S3 + 6H2O ---> 2Al(OH)3 + 3H2S

A) What is the mass of hydrogen sulfide produced in the above reaction?


B) Which is the limiting reagent in the above reaction?

a. aluminum sulfide

b. water

c. hydrogen sulfide

d. aluminum hydroxide

Please show your work so I can see how to do it for future problems! Thank you!

Respuesta :

onyebuchinnaji

(a) The mass of hydrogen sulfide produced in the reaction is 9.44 g.

(b) Water is the limiting reagent in the given reaction.

Mass of hydrogen sulfide produced

Al2S3 + 6H2O ---> 2Al(OH)3 + 3H2S

mass of water in the reaction = 6(18 g/mol) = 108 g/mol

mass of hydrogen sulfide in the reaction: = 3(34) = 102 g/mol

108 g/mol of H2O ------- > 102 g/mol of H2S

10 of H2O -----------> ?

= 9.44 g

Aluminum sulfide will be completely consumed by the reaction while some fraction of water will remain. Thus, water is the limiting reagent in the given reaction.

Learn more about limiting reagents here: https://brainly.com/question/26905271

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