Q9 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Chemistry
For a given unbalanced reaction, $\mathrm{MnO}_2+\mathrm{HCl} \longrightarrow \mathrm{MnCl}_2+\mathrm{H}_2 \mathrm{O}$, which is the limiting reagent, if the initial amount for each of the reactant is 100 grams?
[Molar masses : $\mathrm{MnO}_2=86.9 ; \mathrm{HCl}=36.5$; $\left.\mathrm{MnCl}_2=125.8 ; \mathrm{Cl}_2=70.9 ; \mathrm{H}_2 \mathrm{O}=18\right]$
- A$\mathrm{MnO}_2$
- Correct answerB$\mathrm{HCl}$
- C$\mathrm{MnCl}_2$
- D$\mathrm{Cl}_2$
Solution
Limiting reagent In a chemical reaction, the limiting reagent is the reactant that determine, how much of the products are made. The other reactants are sometimes referred to as being in excess, since there will be some leftover after the limiting reagent is completely used up.
Given,
unbalanced reaction,
$\begin{array}{lcccc} & \mathrm{MnO}_2+ & \mathrm{HCl} & \longrightarrow \mathrm{MnCl}_2+\mathrm{H}_2 \mathrm{O} \\ \text { Initial amount } & 100 \mathrm{~g } & 100 \mathrm{~g} & & \\ \text { of reactant } & & \\ \text { Molar mass } & 86.9 & 36.5 & 125.8 & 18 \end{array}$
Balancing the reaction,
Convert amounts to moles Moles of $\mathrm{MnO}_2=100 \mathrm{~g} \times \frac{1 \mathrm{~mol}}{86.9 \mathrm{~g}}=1.15 \mathrm{~mol}$ of $\mathrm{MnO}_2$ Moles of $\mathrm{HCl}=100 \mathrm{~g} \times \frac{1 \mathrm{~mol}}{36.5 \mathrm{~g}}=2.73 \mathrm{~mol}$ of $\mathrm{HCl}$ Actual ratio (molar ratio) of reactants $=\frac{\text { Moles of } \mathrm{MnO}_2}{\text { Moles of } \mathrm{HCl}}=\frac{1.15 \mathrm{~mol}}{2.73 \mathrm{~mol}}=\frac{0.42}{1} \mathrm{~mol}$ The actual ratio tells us that we have 0.42 moles of $\mathrm{MnO}_2$ for every 1 mole of $\mathrm{HCl}$. In comparison, the stoichiometric ratio from our balanced reaction is below : $\text { Stoichiometric ratio }=\frac{1 \text { moles of } \mathrm{N}_2}{4 \text { moles of } \mathrm{HCl}}$ $=\frac{0.25 \text { moles } \mathrm{MnO}_2}{1 \text { moles } \mathrm{HCl}}$ This means we need at least 0.25 mole of $\mathrm{MnO}_2$ for every mole of $\mathrm{HCl}$. Since, our actual ratio is greater then our stoichiometric ratio, we have more $\mathrm{MnO}_2$ than we need to react with each mole of $\mathrm{HCl}$. Therefore, $\mathrm{HCl}$ is our limiting reagent and $\mathrm{MnO}_2$ is in excess. Hence, correct option is (b).
Convert amounts to moles Moles of $\mathrm{MnO}_2=100 \mathrm{~g} \times \frac{1 \mathrm{~mol}}{86.9 \mathrm{~g}}=1.15 \mathrm{~mol}$ of $\mathrm{MnO}_2$ Moles of $\mathrm{HCl}=100 \mathrm{~g} \times \frac{1 \mathrm{~mol}}{36.5 \mathrm{~g}}=2.73 \mathrm{~mol}$ of $\mathrm{HCl}$ Actual ratio (molar ratio) of reactants $=\frac{\text { Moles of } \mathrm{MnO}_2}{\text { Moles of } \mathrm{HCl}}=\frac{1.15 \mathrm{~mol}}{2.73 \mathrm{~mol}}=\frac{0.42}{1} \mathrm{~mol}$ The actual ratio tells us that we have 0.42 moles of $\mathrm{MnO}_2$ for every 1 mole of $\mathrm{HCl}$. In comparison, the stoichiometric ratio from our balanced reaction is below : $\text { Stoichiometric ratio }=\frac{1 \text { moles of } \mathrm{N}_2}{4 \text { moles of } \mathrm{HCl}}$ $=\frac{0.25 \text { moles } \mathrm{MnO}_2}{1 \text { moles } \mathrm{HCl}}$ This means we need at least 0.25 mole of $\mathrm{MnO}_2$ for every mole of $\mathrm{HCl}$. Since, our actual ratio is greater then our stoichiometric ratio, we have more $\mathrm{MnO}_2$ than we need to react with each mole of $\mathrm{HCl}$. Therefore, $\mathrm{HCl}$ is our limiting reagent and $\mathrm{MnO}_2$ is in excess. Hence, correct option is (b).
Hence, the correct answer is option: 'B'.