Question

A metal oxide has the formula ${X_2}{O_3}$. It can be reduced by hydrogen to give free metal and water. 0.1596g of metal oxide requires 6mg of Hydrogen for complete reduction. Calculate the atomic weight of the metal (in amu)

Answer 1

Hint: By applying the law of conservation of mass to a balanced reaction, we can say that the mass of components on the reactant side will have to be the same as the mass of the products.
Hence, We can use this property to determine the mass of metal in a molecule.

Complete step by step answer:
Firstly we have to understand what a metal oxide is.
Both metals and nonmetals can attain their highest oxidation state in compounds with oxygen. The alkali metals and the alkaline earth metals, as well as the transition metal and the post transition metals from ionic oxides that are compounds that contain the ${O^{2 - }}$ anion. Metal with high oxidation states form oxides whose bonds have a more covalent nature. Nonmetals also form covalent oxide, which are usually molecular in character.
Metal oxide are crystalline solids that contain a metal cation and an oxide anion. They typically react with water to form bases or with acids to form salts. These are of two type,
Basic oxides: these are those which react with water to give base. Examples include the oxide of most metals.
Acidic oxide: these are those which react with water and give acid. They are oxide of either nonmetals or metals in highly oxidation states.
Amphoteric oxide: it is an oxide that can act as either acid or base in a reaction to produce a salt and water. Amphotericity depends on the oxidation states available to a chemical species. Because metals have multiple oxidation states, they form amphoteric oxide. Examples are $A{l_2}{O_3},ZnO$
Now, comes to the solution part when Metal oxide reduced is by hydrogen
the reaction can be written as,
${X_2}{O_3} + 3{H_2} \to 2X + 3{H_2}O$
$0.006g$ Of ${H_2}$ is required by $0.159g$ of oxide
$6g$ of ${H_2}$ Will be required by $159.6g$ oxide
Then, atomic mass of ${X_2}{O_3}$ is calculated as
$2(X) + 3(O) \Rightarrow 2(X) + 3(16) = 159.6$
Solving the above equation we get:
$2X + 48 = 159.6$
On further simplification we get:
$\dfrac{{159.6 - 48}}{2}$
Hence, we get:
$X = 55.8g$
Hence, the required answer is $55.8g$.

Note:Metal oxides are extreme hydrolysis products of metals with water. Metals which are non-acidic exist as only their oxide forms such as sodium metal react irreversibly with water to form sodium oxide, however, metals which are very strongly acidic such as tellurium exist as hydroxide for a large $pH$ range. The stability of metal oxide with respect to the $pH$ of the solution is given by a predominance diagram.