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