Question

A 3.540-g sample of an unknown metal M is burned in the presence of excess oxygen, producing the oxide M2O3(s) and liberating 18.56 kJ of heat at constant pressure. What is the identity of the metal? 4M(s) + 3O2(g) → 2M2O3(s) SubstanceΔH°f (kJ/mol) Yb2O3(s)–1814.6 Tb2O3(s)–1865.2 Sm2O3(s)–1823.0 Sc2O3(s)–1908.8 Y2O3(s)–1905.3 a) Sm b) Tb c) Y d) Sc e) Yb

Answer 1

Answer:

a) Sm

Explanation:

Hello,

This substantiation could be done by knowing how to compute the released energy for 1 mol the oxide. Thus, for Sm, the computation turns out into:

$Q=-1823.0\frac{kJ}{mol}*\frac{1molSm_2O_3}{348.72gSm_2O_3} *3.540gSm_2O_3\\Q=-18.5kJ$

The sign just indicates that the heat is released due to the chemical reaction and it matches with the indicated heat in the statement.

Best regards.