Question

How many grams of magnesium is required to dissolve in dilute ${{H}_{2}}S{{O}_{4}}$, so that the liberated ${{H}_{2}}$, gas is just sufficient to reduce 160 grams of ferric oxide?
a.) 24
b.) 48
c.) 73
d.) 96

Answer 1

To solve this question, first we have to write the balanced chemical reaction which is involved. After that we will use the stoichiometric method to calculate the answer.

Complete Solution : Given in the question,
Mass of ferric oxide = 160 grams
The balanced chemical reaction involved:
$Mg+{{H}_{2}}S{{O}_{4}}\to MgS{{O}_{4}}+{{H}_{2}}F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\to 2Fe+3{{H}_{2}}O$
Atomic mass of oxygen is = 16 g/mol
Atomic mass of magnesium = 24.3 g/mol
Atomic mass of iron = 55.8 g/mol
Assuming that to reduce to 160g of ferric oxide the amount of magnesium which is required be w grams.
$=\dfrac{160}{2\times 55.8+3\times 16}mol~F{{e}_{2}}{{O}_{3}}\times \dfrac{3\text{ mol Mg}}{1\text{ mol F}{{\text{e}}_{2}}{{O}_{3}}}=\dfrac{w}{24.3}mol~Mg$
Hence, the amount of magnesium which is required

& w=\dfrac{160}{2\times 55.8 + 3\times 16}\times 3\times 24.3 \\\ & w=73g \\\ \end{aligned}$$ Hence, the correct answer is option (C) i.e. 74 grams of magnesium is required to dissolve in dilute ${{H}_{2}}S{{O}_{4}}$, so that the liberated ${{H}_{2}}$ gas is just sufficient to reduce 160 grams of ferric oxide. **So, the correct answer is “Option C”.** **Additional information** - To calculate the number of atoms first we have to calculate the number of moles of the given element as the number of moles is already given, then the number of atoms will be equal to the product of number of moles and Avogadro’s number. A mole is generally used as a unit for the number of atoms or molecules or particles of a material. **Note:** While solving this type of questions, first we have to balance the chemical reaction. If the chemical reaction is not balanced then the stoichiometric calculations will never be correct.