Answer
a) $30\ mL$
b) $0.627\ g$
Work Step by Step
a) The number of moles of oxalic acid:
$28\ g\times 1.2/100\div90.03\ g/mol=0.00373\ mol$
Oxalic acid is a diprotic acid, so two moles of NaOH is required to neutralize each mole of oxalic acid.
$0.25\ M\times V = 2\times0.00373\ mol$
$V=0.030\ L = 30\ mL$
b) 1 mol of calcium oxalate for each mol of oxalic acid:
$0.00373\ mol\times 168.18\ g/mol=0.627\ g$
