Question

For a given unbalanced reaction, $MnO _{2}+ HCl \rightarrow MnCl _{2}+ H _{2} O$, which is the limiting reagent if the initial amount for each of the reactant is 100 grams?
[Molar masses: $MnO _{2}=86.9 ; HCl =36.5 ; MnCl _{2}=125.8; Cl _{2}=70.9 ; H _{2} O =18$ ]

• A. $MnO _{2}$
• B. $HCl$
• C. $MnCl _{2}$
• D. $Cl _{2}$

Answer 1

Answer: (B)

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,

Convert amounts to moles
Moles of $MnO _{2}=100 \,g \times \frac{1 \,mol }{86.9 \,g }=1.15 \,mol$ of $MnO _{2}$
Moles of $HCl =100\, g \times \frac{1 \,mol }{36.5 \,g }=2.73 \,mol$ of $HCl$
Actual ratio (molar ratio) of reactants
$=\frac{\text { Moles of } MnO _{2}}{\text { Moles of } HCl }=\frac{1.15\, mol }{2.73\, mol }=\frac{0.42}{1} mol$
The actual ratio tells us that we have $0.42$ moles of $MnO _{2}$ for every $1$ mole of $HCl$. In comparison, the stoichiometric ratio from our balanced reaction is below:
Stoichiometric ratio $=\frac{1 \text { moles of } N_{2}}{4 \text { moles of } H C l}$
$=\frac{0.25 \text{moles} MnO _{2}}{1 \text{moles} HCl }$
This means we need at least $0.25$ mole of $MnO _{2}$ for every mole of $HCl$. Since, our actual ratio is greater then our stoichiometric ratio, we have more $MnO _{2}$ than we need to react with each mole of $HCl$.
Therefore, $HCl$ is our limiting reagent and $MnO _{2}$ is in excess.