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Question

316.0 g of aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?
The unbalanced equation is
Al2S3+H2OAl(OH)3+H2S
  1. 312.56 g
  2. 123.89 g
  3. 400.36 g
  4. 265.14 g


Solution

The correct option is D 265.14 g
Al2S3+6H2O2Al(OH)3+3H2S
Determining moles of the reactant:
For Al2S3316150=2.11 mol
For H2O49318=27.39 mol
Finding limiting reagent
For Al2S3=2.111=2.11
For H2O=27.396=4.56
Al2S3 is the limiting reagent.
1 mol of Al2S3 requires 6 mol of water
2.11 mol of Al2S3 will require =6×2.11=12.66 mol
Mass of water consumed =12.66×18=227.88 g
Mass of excess reactant left =493227.88=265.12 g

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