Consider the following Chemical equation: $aAl_2O_3 + bHCl \longrightarrow cAlCl_3 + dH_2O$. In order to balance this chemical equation, the values of a, b, c and d must be

To balance the chemical equation:

$$aAl_2O_3 + bHCl \longrightarrow cAlCl_3 + dH_2O$$

we count the atoms of each element on both sides of the equation:

1. Aluminum (Al): There are $2a$ atoms on the LHS and $c$ atoms on the RHS. Thus, $2a = c$.
2. Oxygen (O): There are $3a$ atoms on the LHS and $d$ atoms on the RHS. Thus, $3a = d$.
3. Hydrogen (H): There are $b$ atoms on the LHS and $2d$ atoms on the RHS. Thus, $b = 2d$.
4. Chlorine (Cl): There are $b$ atoms on the LHS and $3c$ atoms on the RHS. Thus, $b = 3c$.

Let us assume $a = 1$:

• From $2a = c$, we get $c = 2(1) = 2$.
• From $3a = d$, we get $d = 3(1) = 3$.
• From $b = 2d$, we get $b = 2(3) = 6$.

Let's verify with the chlorine atoms: $b = 6$ on the LHS and $3c = 3(2) = 6$ on the RHS, which is perfectly balanced.

Thus, the coefficients are $a = 1$, $b = 6$, $c = 2$, and $d = 3$. This corresponds to Option A.

• Options B, C, and D are incorrect because they do not satisfy the conservation of atoms for all elements simultaneously.