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Question

Suppose 316.0 g aluminium sulphide reacts with 493.0 g of water. How much of the excess reactant is left over at the end of the reaction?
Al2S3+H2OAl(OH)3+H2S
 
  1. 265.4 g
  2. 309.0 g
  3. 109.8 g
  4. 340.3 g


Solution

The correct option is A 265.4 g
The unbalanced equation is:
Al2S3+H2OAl(OH)3+H2S
The balanced equation is: 
Al2S3+6H2O2Al(OH)3+3H2S
Number of moles of Al2S3=316.0 g150 g/mol=2.107 mol
Number of moles of  H2O=493.0 g18g/mol=27.388 mol

For Al2S3 
Initial number of molesstoichiometric coefficient=2.1071=2.1044

For  H2O  Initial number of molesstoichiometric coefficient=27.3886=4.564

Al2S3 is the limiting reagent.

To find the number of moles of  H2O required to consume 2.107 mol of Al2S3:

(2.107 mol ofAl2S3×6 mol of H2O)(1 mol of Al2S3)=12.642 mol of H2O

Weight of H2O=12.642 mol×18 g/mol=227.556 g

Number of moles of excess reagent left = 493.0 g – 227.556 = 265.444 g

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