Question

metal oxide having formula \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] gives free metal and water on reduction with hydrogen. $0 \cdot 2{\text{ g}}$ of metal oxide can be completely reduced by $12{\text{ mg}}$ of hydrogen. The atomic weight of the metal is:
A. ${\text{12}}$
B. ${\text{13}}$
C. ${\text{26}}$
D. ${\text{78}}$

Answer 1

Hint: The atomic weight of any element is the product of the equivalent weight of the metal and its valency. Thus, calculate the equivalent weight and valency of the metal. The equivalent weight can be calculated from the mass of metal oxide reduced by one gram of hydrogen. The valency can be calculated from the formula of the metal oxide.

Complete step by step answer:
Step 1: Write the balanced chemical equation as follows:
A metal oxide having formula \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] gives free metal and water on reduction with hydrogen. Thus, the reaction is,
\[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {{\text{H}}_2} \to {\text{Z}} + {{\text{H}}_2}{\text{O}}\]
Count the number of all the atoms on the reactant and product sides. Thus,
\[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {{\text{H}}_2} \to {\text{Z}} + {{\text{H}}_2}{\text{O}}\]
\[{\text{ Z}} - 2{\text{ Z}} - 1\]
\[{\text{ O}} - 3{\text{ O}} - 1\]
\[{\text{ H}} - 2{\text{ H}} - 2\]
Change the coefficient of \[{\text{Z}}\] to $2$ to balance the number of \[{\text{Z}}\] atoms. Thus,
\[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {{\text{H}}_2} \to 2{\text{Z}} + {{\text{H}}_2}{\text{O}}\]
\[{\text{ Z}} - 2{\text{ Z}} - 2\]
\[{\text{ O}} - 3{\text{ O}} - 1\]
\[{\text{ H}} - 2{\text{ H}} - 2\]
Change the coefficient of \[{{\text{H}}_2}{\text{O}}\] to $3$ to balance the number of ${\text{O}}$ atoms. Thus,
\[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {{\text{H}}_2} \to 2{\text{Z}} + 3{{\text{H}}_2}{\text{O}}\]
\[{\text{ Z}} - 2{\text{ Z}} - 2\]
\[{\text{ O}} - 3{\text{ O}} - 3\]
\[{\text{ H}} - 2{\text{ H}} - 6\]
Change the coefficient of \[{{\text{H}}_2}\] to $3$ to balance the number of ${\text{H}}$ atoms. Thus,
\[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 3{{\text{H}}_2} \to 2{\text{Z}} + 3{{\text{H}}_2}{\text{O}}\]
\[{\text{ Z}} - 2{\text{ Z}} - 2\]
\[{\text{ O}} - 3{\text{ O}} - 3\]
\[{\text{ H}} - 6{\text{ H}} - 6\]
Thus, the balanced chemical equation is,
\[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 3{{\text{H}}_2} \to 2{\text{Z}} + 3{{\text{H}}_2}{\text{O}}\]
Step 2: Convert the units of weight of hydrogen from ${\text{mg}}$ to ${\text{g}}$ using the relation as follows:
$1{\text{ mg}} = 1 \times {10^{ - 3}}{\text{ g}}$
Thus,
${\text{Mass of hydrogen}} = 12{\text{ }}\not{{{\text{mg}}}} \times \dfrac{{1 \times {{10}^{ - 3}}{\text{ g}}}}{{1{\text{ }}\not{{{\text{mg}}}}}}$
${\text{Mass of hydrogen}} = 12 \times {10^{ - 3}}{\text{ g}}$
Thus, the weight of hydrogen is $12 \times {10^{ - 3}}{\text{ g}}$.
Step 3: Calculate the weight of \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] reduced by $1{\text{ g}}$ of hydrogen as follows:
$0 \cdot 2{\text{ g}}$ of \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] is reduced by $12 \times {10^{ - 3}}{\text{ g}}$. Thus, weight of \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] reduced by $1{\text{ g}}$ of hydrogen is,
${\text{Mass of }}{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 1{\text{ }}\not{{{\text{g }}{{\text{H}}_2}}} \times \dfrac{{0 \cdot 2{\text{ g }}{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}}}{{12 \times {{10}^{ - 3}}{\text{ }}\not{{{\text{g }}{{\text{H}}_2}}}}}$
${\text{Mass of }}{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 16 \cdot 7{\text{ g}}$
Thus, the weight of \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] reduced by $1{\text{ g}}$ of hydrogen is $16 \cdot 7{\text{ g}}$.
Step 4: Calculate the equivalent weight of the metal as follows:
The equivalent weight of the metal oxide is the sum of the equivalent weights of the metal and oxygen. Thus,
${\text{Equivalent weight of }}{{\text{Z}}_2}{{\text{O}}_3} = {\text{Equivalent weight of Z}} + {\text{Equivalent weight of O}}$
Rearrange the equation for the equivalent weight of the metal. Thus,
${\text{Equivalent weight of }}{{\text{Z}}_2}{{\text{O}}_3} = {\text{Equivalent weight of Z}} + {\text{Equivalent weight of O}}$
${\text{Equivalent weight of Z}} = {\text{Equivalent weight of }}{{\text{Z}}_2}{{\text{O}}_3} - {\text{Equivalent weight of O}}$
Substitute $16 \cdot 7$ for the equivalent weight of \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\], $8$ for the equivalent weight of oxygen. Thus,
${\text{Equivalent weight of Z}} = 16 \cdot 7 - {\text{8}}$
${\text{Equivalent weight of Z}} = 8 \cdot 7$
Thus, the equivalent weight of the metal is $8 \cdot 7$.
Step 5: Calculate the atomic weight of the metal as follows:
The atomic weight of any element is the product of the equivalent weight of the metal and its valency. Thus,
${\text{Atomic weight of Z}} = {\text{Equivalent weight of Z}} \times {\text{Valency of Z in }}{{\text{Z}}_2}{{\text{O}}_3}$
The metal oxide is \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]. The valency of the metal in \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] is $3$.
Substitute $8 \cdot 7$ for the equivalent weight of ${\text{Z}}$, $3$ for the valency of ${\text{Z}}$. Thus,
${\text{Atomic weight of Z}} = 8 \cdot 7 \times 3$
${\text{Atomic weight of Z}} = 26$
Thus, the atomic weight of the metal is ${\text{26}}$.

So, the correct answer is Option c.

Note:
The combining power of any metal with other atoms to form a chemical compound is known as the valency of the metal. Thus,
For \[{{\text{Z}}_{\text{2}}}{{\text{O}}_{\text{3}}}\], two atoms of ${\text{Z}}$ combine with three atoms of ${\text{O}}$. Thus, the valency of ${\text{Z}}$ is three and that of ${\text{O}}$ is two.