Balanced chemical equation for reaction of aluminium hydroxide and sulfuric acid is as follows:
![2Al(OH)_{3}(s)+3H_{2}SO_{4}(aq)\rightarrow Al_{2}(SO_{4})_{3}(aq)+6H_{2}O(l)](https://tex.z-dn.net/?f=2Al%28OH%29_%7B3%7D%28s%29%2B3H_%7B2%7DSO_%7B4%7D%28aq%29%5Crightarrow%20Al_%7B2%7D%28SO_%7B4%7D%29_%7B3%7D%28aq%29%2B6H_%7B2%7DO%28l%29)
(a) From the balanced chemical reaction, 2 mole of
gives 1 mole of
thus, 1 mole of
gives 0.5 moles of
and 0.410 moles of
gives 0.205 moles of
.
Molar mass of
is 342.1509 g/mol thus, mass of
will be:
m=n×M=0.205 mol×342.1509 g/mol=70.14 g
Similarly, 3 moles of
gives 1 mole of
thus, 1 mole of
gives 0.33 moles of
and 0.410 moles of
gives 0.136 moles of
.
Thus, mass of
will be:
m=n×M=0.136 mol×342.1509 g/mol=46.76g
Since, mass of
obtained from
is less than that from
,
is the limiting reactant.
(b) Number of moles of
is 0.410 mol, from the balanced chemical reaction 3 moles of
gives 1 mole of
thus, 1 mole of
gives 0.33 moles of
and 0.410 moles of
gives 0.136 moles of
.
Therefore, number of moles of
formed is 0.136 mol.
(c) After the completion of reaction, all of the limiting reactant gets converted into product thus, 0.410 mole of
reacts completely to give 0.136 mol of
.
From the balanced chemical reaction, 1 mole of
formed from 2 moles of
, thus, 0.136 mol of
formed from 0.136×2=0.272 mol.
Thus, 0.272 moles of
is used and remaining moles will be (0.410-0.272) mol=0.138 mol.
Therefore, 0.138 mol of excess reactant remains after the completion of reaction.