For the following reaction if equal mass of A and B are taken $\mathrm{A}+2\mathrm{B}\to \mathrm{C}$
Which of the following is/are correct. [ ${\mathrm{M}}_{\mathrm{A}}$ and ${\mathrm{M}}_{\mathrm{B}}$ are the molar masses of A and B respectively]

According to the given reaction, the mole ratio of A and B is 1:2. We can Assume that mass of A and B is 1 g each and molar mass of A is 1 g/mol.

$$\begin{gathered} \mathrm{moles}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}} \\ {\mathrm{Moles}}_{\mathrm{A}}=\frac{{\mathrm{mass}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{A}}} \\ \frac{1}{1 \mathrm{g}/\mathrm{mol}}({\mathrm{M}}_{\mathrm{A}}=1 \mathrm{g}/\mathrm{mol})=1 \mathrm{mole} \\ {\mathrm{Moles}}_{\mathrm{B}}=\frac{\mathrm{massB}}{{\mathrm{M}}_{\mathrm{B}}} \\ \frac{1}{{\displaystyle \frac{{\mathrm{M}}_{\mathrm{A}}}{2}}}=2 \mathrm{mole} (\mathrm{using} {\mathrm{M}}_{\mathrm{A}}=2{\mathrm{M}}_{\mathrm{B}}) \end{gathered}$$

Mole ratio of A and B is 1:2, so all reactant is consumed.

Assume that A has molar mass=1 g/mol and B is 0.1 g/mol.

$$\begin{gathered} {\mathrm{Moles}}_{\mathrm{A}}=\frac{{\mathrm{mass}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{A}}} \\ \frac{1}{1 \mathrm{g}/\mathrm{mol}}({\mathrm{M}}_{\mathrm{A}}=1 \mathrm{g}/\mathrm{mol})=1 \mathrm{mole} \\ {\mathrm{Moles}}_{\mathrm{B}}=\frac{\mathrm{massB}}{{\mathrm{M}}_{\mathrm{B}}} \\ \frac{1}{{\displaystyle 0.1}}=10 \mathrm{mole} \end{gathered}$$

As ${\mathrm{M}}_{\mathrm{B}}<\frac{{\mathrm{M}}_{\mathrm{A}}}{2}$, A will be limiting reagent.