Answer
$m(H_2TeO_3) = 82.5475g$
Work Step by Step
Considering stoichiometric coefficients, we can conclude that 1 mole of $TeO_2$ produces 1 mole of ${TeO_{3}}^{2-}$, and 1 mole of ${TeO_{3}}^{2-}$ produces 1 mole of $H_2TeO_3$. Hence, 1 mole of $TeO_2$ produces 1 mole of ${TeO_{3}}^{2-}$.
$n(H_2TeO_3) = n(TeO_2) = \frac{m(TeO_2)}{M(TeO_2)}=\frac{74.2g}{160\frac{g}{mol}}=0.46375mol\implies m(H_2TeO_3) = M(H_2TeO_3)\times n(H_2TeO_3) = 178\frac{g}{mol}\times 0.46375mol=82.5475g$
