Chapter 4, Problem 4.151MP

Element M is prepared industrially by a two-step procedure according to the following (unbalanced) equations:
(1) ${\text{M}}_2{\text{O}}_3\left(s\right)+\text{C}\left(s\right)+{\text{Cl}}_2\left(g\right)\to {\text{MCl}}_3\left(l\right)+\text{CO}\left(g\right)$
(2) ${\text{MCl}}_3\left(l\right)+{\text{H}}_{\text{2}}\left(g\right)\to \text{M}\left(s\right)+\text{HCl}\left(g\right)$
Assume that 0.855 g of ${\text{M}}_{\text{2}}{\text{O}}_{\text{3}}$ is submitted to the reaction sequence. When the HCI produced in step (2) is dissolved in water and titrated with 0.511 M NaOH, 144.2 mL of theNaOH solution is required to neutralize the HCl.
(a) Balance both equations.
(b) What as the atomic mass of element M, and what is its identity?
(c) What mass of M in grams is produced in the reaction?