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

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.