Elements A and B react completely to form a compound with the...

The balanced chemical equation for the formation of the compound is:

$2\text{A} + 3\text{B} \rightarrow \text{A}_2\text{B}_3$

From the stoichiometry of the reaction, $2$ moles of $\text{A}$ react with $3$ moles of $\text{B}$.

Therefore, the ratio of moles of $\text{A}$ to moles of $\text{B}$ reacting is $\dfrac{n_{\text{A}}}{n_{\text{B}}} = \dfrac{2}{3}$.

Let the atomic masses of $\text{A}$ and $\text{B}$ be $M_{\text{A}}$ and $M_{\text{B}}$ respectively.

Number of moles of $\text{A}$, $n_{\text{A}} = \dfrac{40}{M_{\text{A}}}$.

Number of moles of $\text{B}$, $n_{\text{B}} = \dfrac{10}{M_{\text{B}}}$.

Substituting these into the mole ratio:

$\dfrac{\dfrac{40}{M_{\text{A}}}}{\dfrac{10}{M_{\text{B}}}} = \dfrac{2}{3}$

$4 \times \dfrac{M_{\text{B}}}{M_{\text{A}}} = \dfrac{2}{3}$

Rearranging the equation to find the ratio of atomic masses:

$\dfrac{M_{\text{A}}}{M_{\text{B}}} = \dfrac{4 \times 3}{2} = 6$.

Answer: $6$