Suppose 316.0 g of aluminium sulphide reacts with 493.0 g of water, what mass of the excess reactant will remain?
Given reaction: Al2S3+6H2O→2Al(OH)3+3H2S
A
265.14 g
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B
108.52 g
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C
400 g
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D
66.25 g
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Solution
The correct option is A 265.14 g The unbalanced equation is: Al2S3+H2O→Al(OH)3+H2S
On balancing the equation, Al2S3+6H2O→2Al(OH)3+3H2S
Moles of Al2S3=316150 =2.11 moles
Moles of water =49318=27.39 moles
1 mole of Al2S3 reacts with 6 moles of water.
2.11 moles of Al2S3 reacts with 61×2.11=12.66 moles of water
So, Al2S3 is the limiting reagent.
Mass of excess reagent left = (27.39−12.66) mole ×18gmole
Mass of excess reagent left = 265.14 g