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3.0 g of H(2) react with 29.0 g O(2) to ...

3.0 g of `H_(2)` react with 29.0 g `O_(2)` to yield `H_(2)O`
(i) What is the limiting reactant ?
(ii) Calculate the maximum amount of water that can be formed
(iii) Calculate the amount of one of the reactants which remains unreacted.

Text Solution

Verified by Experts

The balanced chemical equation is
`underset(2"mol")(2H_(2)) + underset("1 mol")(O_(2)) to underset(2 "mol")(2H_(2)O)`
Moles of `H_(2) =("Wt. of" H_(2))/("Molecular mass of" H_(2)) = 3.0/2=1.50` mol
moles of `O_(2) = ("Wt. of"O_(2))/("Molecular mass of" O_(2)) = (22.0)/(32.0)`
`=0.906` mol
(i) Calculation of limiting reactant
According to the above equation, 2 mol of `H_2` require `O_(2)` = 1 mol 1
1.50 mol of `H_2` require `O_(2) =1.50/2 = 0.75` mol
But moles of `O_(2)` actually present = 0.906 mol
Therefore, `O_(2)` is in excess and `H_(2)` is the limiting reactant.
(it) Calculation of maximum amount of `H_(2)O` formed.
Now, 2 mol of `H_(2)` form`H_(2)O` = 2 mol
1.50 mol of `H_(2)` form `H_(2)O`= 1.50 mol
Thus, the maximum amount of `H_(2)O` formed = 1.50 mol
(iii) Calculation of the amount of one of the reactants which remains unreacted i.e.`O_(2)`
Number of moles of `O_(2)` added = 0.906 mol
Number of moles of `O_(2)` used up = 0.75 mol Number of moles of`O_(2)` unreacted = 0.906 - 0.75 = 0.156 mol.
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