Given:
The mass of NaOH = 165 g.
The molar mass of NaOH = 40 g/mol.
The moles of NaOH = 165 g/40 g/mol = 4.125 mol.
According to the stoichiometry of the reaction
6NaOH+2Al---------->2Na3AlO3 +3H2.
6 mol of NaOH produces = 2 mol of Na3AlO3.
1 mol of NaOH produces = (1/3) mol of Na3AlO3.
4.125 mol of NaOH produces = {(1/3)*4.125} mol of Na3AlO3.
= 1.375 mol of Na3AlO3.
Now,
mass of Na3AlO3 = mol of Na3AlO3 * molar mass of Na3AlO3.
=1.375*143.94
=197.9175 g. {Answer}