Answer
See answer below.
Work Step by Step
a) Sodium iodate, sodium bisulfite
b) Number of moles of iodine:
$1000\ g\div 253.81\ g/mol=3.94\ mol$
Mass of iodate:
$3.94\ mol\times 2/1 \times 196.89\ g/mol=1551.5\ g$
Mass of bisulfite:
$3.94\ mol\times 5/1 \times 104.06\ g/mol=2049.9\ g$
c) Number of moles of iodate:
$15.0\ g\div 196.89\ g/mol=0.0762\ mol$
Number of moles of bisulfite:
$0.125\ L\times 0.853\ M=0.107\ mol$
Ratio: $1.4\gt 5/2$, iodate is the limiting reactant.
Mass of iodine:
$0.0762\ mol\times 1/2\times 253.81\ g/mol=9.67\ g$
