Question

How many grams of \[{\text{Al}}\]are required to completely react with \[81.2{\text{ g}}\]of \[{\text{Mn}}{{\text{O}}_2}\]? What is the mole ratio between \[{\text{Mn}}{{\text{O}}_2}\]and \[{\text{Al}}\]in the balanced equation?
\[{\text{3}}\,{\text{Mn}}{{\text{O}}_2} + 4\,{\text{Al}} \to {\text{3}}\,{\text{Mn + 2}}\,{\text{A}}{{\text{l}}_2}{{\text{O}}_3}\]

Answer 1

Hint: The first part of question can be solved by applying mass to mass stoichiometry between \[{\text{Mn}}{{\text{O}}_2}\] and \[{\text{Al}}\]. If we look into the given balanced chemical equation, the second part of the question can be easily answered.

Complete step by step answer:
Molar mass of \[{\text{Mn}} = 55{\text{ g mo}}{{\text{l}}^{ - 1}}\] and that of \[{\text{O}} = 16{\text{ g mo}}{{\text{l}}^{ - 1}}\]
Molar mass of \[{\text{Mn}}{{\text{O}}_2} = (1 \times 55) + (2 \times 16) = 87{\text{ g mo}}{{\text{l}}^{ - 1}}\]
Moles of \[{\text{Mn}}{{\text{O}}_2} = \dfrac{{{\text{Mass of Mn}}{{\text{O}}_2}}}{{{\text{Molar mass}}}}\]
\[ = \dfrac{{81.2}}{{87}}{\text{ mol}}\]
Molar mass of \[{\text{Al}} = 27\]
The given balanced equation is:
\[{\text{3}}\,{\text{Mn}}{{\text{O}}_2} + 4\,{\text{Al}} \to {\text{3}}\,{\text{Mn + 2}}\,{\text{A}}{{\text{l}}_2}{{\text{O}}_3}\]
Applying mass to mass stoichiometry:
\[\because 3\]moles of \[{\text{Mn}}{{\text{O}}_2}\]react completely with \[ = 4\]moles of \[{\text{Al}}\]
\[\therefore 1\]mole of \[{\text{Mn}}{{\text{O}}_2}\]react completely with \[ = \dfrac{4}{3}\]moles of \[{\text{Al}}\]
\[\therefore \dfrac{{81.2}}{{87}}\]moles of \[{\text{Mn}}{{\text{O}}_2}\]react completely with \[ = \dfrac{4}{3} \times \dfrac{{81.2}}{{87}}\] moles of \[{\text{Al}}\]
Upon solving the above expression, moles of \[{\text{Al}} = \dfrac{{36}}{{29}} = 1.24{\text{ mol}}\]
Mass of \[{\text{Al}} = \]Number of moles \[ \times \] molar mass
\[ = 1.24 \times 27{\text{ g}}\]
\[ = 33.48{\text{ g}}\]
When we round this to three significant figures, we get:
Required mass of aluminium\[ = 33.5{\text{ g}}\]
The asked mole ratio will be the ratio of stoichiometric coefficients in the balanced chemical equation. We can see that \[3\] moles of \[{\text{Mn}}{{\text{O}}_2}\] react with \[4\] moles of \[{\text{Al}}\].

Hence, the mole ratio of \[{\text{Mn}}{{\text{O}}_2}\] and \[{\text{Al}}\]is \[3:4\].

Note: In this solution, moles of aluminium are calculated by using ‘unitary method’. This method is a technique for solving a mathematical problem by first finding the value of a single unit and then finding the required value by multiplying this value of a single unit.
In this method, first we will calculate the value for \[1\] mole of \[{\text{Mn}}{{\text{O}}_2}\] from the known value for \[3\]moles of \[{\text{Mn}}{{\text{O}}_2}\]. Then we multiplied the obtained value with the required multiple to get our value.
We should remember the molar mass of oxygen, aluminium and manganese to solve this problem.