Answer :
To balance the chemical equation:
[tex]\[ a \, \mathrm{Al_2O_3} + b \, \mathrm{HCl} \longrightarrow c \, \mathrm{AlCl_3} + d \, \mathrm{H_2O} \][/tex]
we must determine the correct coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] such that the number of atoms for each element is equal on both sides of the equation.
Starting with the unbalanced equation:
[tex]\[ \mathrm{Al_2O_3} + \mathrm{HCl} \longrightarrow \mathrm{AlCl_3} + \mathrm{H_2O} \][/tex]
Let's follow a step-by-step process to balance it:
### 1. Balance Aluminum (Al):
On the left side, we have 2 Al atoms in [tex]\(\mathrm{Al_2O_3}\)[/tex], so we need the same number of Al atoms on the right side. Since [tex]\(\mathrm{AlCl_3}\)[/tex] contains 1 Al atom, we need 2 moles of [tex]\(\mathrm{AlCl_3}\)[/tex]:
[tex]\[ a \, \mathrm{Al_2O_3} + b \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + d \, \mathrm{H_2O} \][/tex]
### 2. Balance Chlorine (Cl):
Now, we have 2 moles of [tex]\(\mathrm{AlCl_3}\)[/tex], and each mole of [tex]\(\mathrm{AlCl_3}\)[/tex] contains 3 Cl atoms, giving us a total of [tex]\(2 \times 3 = 6\)[/tex] Cl atoms on the right side. Thus, to balance the Cl atoms, we need 6 moles of HCl on the left side:
[tex]\[ a \, \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + d \, \mathrm{H_2O} \][/tex]
### 3. Balance Hydrogen (H):
Each mole of HCl contains 1 H atom, giving us 6 H atoms on the left side. On the right side, each mole of [tex]\(\mathrm{H_2O}\)[/tex] contains 2 H atoms. Therefore, we need 3 moles of [tex]\(\mathrm{H_2O}\)[/tex] to balance the 6 H atoms:
[tex]\[ a \, \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + 3 \, \mathrm{H_2O} \][/tex]
### 4. Balance Oxygen (O):
Finally, we need to balance the O atoms. On the left side, we have 3 O atoms in [tex]\(\mathrm{Al_2O_3}\)[/tex]. On the right side, we have 3 O atoms in [tex]\(3 \, \mathrm{H_2O}\)[/tex]. This matches, so the O atoms are balanced:
[tex]\[ \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + 3 \, \mathrm{H_2O} \][/tex]
Thus, the balanced equation is:
[tex]\[ \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \mathrm{AlCl_3} + 3 \mathrm{H_2O} \][/tex]
Therefore, the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] are:
[tex]\[ (a, b, c, d) = (1, 6, 2, 3) \][/tex]
This matches with option (A).
So, the correct answer is:
(A) [tex]\(1, 6, 2, 3\)[/tex]
[tex]\[ a \, \mathrm{Al_2O_3} + b \, \mathrm{HCl} \longrightarrow c \, \mathrm{AlCl_3} + d \, \mathrm{H_2O} \][/tex]
we must determine the correct coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] such that the number of atoms for each element is equal on both sides of the equation.
Starting with the unbalanced equation:
[tex]\[ \mathrm{Al_2O_3} + \mathrm{HCl} \longrightarrow \mathrm{AlCl_3} + \mathrm{H_2O} \][/tex]
Let's follow a step-by-step process to balance it:
### 1. Balance Aluminum (Al):
On the left side, we have 2 Al atoms in [tex]\(\mathrm{Al_2O_3}\)[/tex], so we need the same number of Al atoms on the right side. Since [tex]\(\mathrm{AlCl_3}\)[/tex] contains 1 Al atom, we need 2 moles of [tex]\(\mathrm{AlCl_3}\)[/tex]:
[tex]\[ a \, \mathrm{Al_2O_3} + b \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + d \, \mathrm{H_2O} \][/tex]
### 2. Balance Chlorine (Cl):
Now, we have 2 moles of [tex]\(\mathrm{AlCl_3}\)[/tex], and each mole of [tex]\(\mathrm{AlCl_3}\)[/tex] contains 3 Cl atoms, giving us a total of [tex]\(2 \times 3 = 6\)[/tex] Cl atoms on the right side. Thus, to balance the Cl atoms, we need 6 moles of HCl on the left side:
[tex]\[ a \, \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + d \, \mathrm{H_2O} \][/tex]
### 3. Balance Hydrogen (H):
Each mole of HCl contains 1 H atom, giving us 6 H atoms on the left side. On the right side, each mole of [tex]\(\mathrm{H_2O}\)[/tex] contains 2 H atoms. Therefore, we need 3 moles of [tex]\(\mathrm{H_2O}\)[/tex] to balance the 6 H atoms:
[tex]\[ a \, \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + 3 \, \mathrm{H_2O} \][/tex]
### 4. Balance Oxygen (O):
Finally, we need to balance the O atoms. On the left side, we have 3 O atoms in [tex]\(\mathrm{Al_2O_3}\)[/tex]. On the right side, we have 3 O atoms in [tex]\(3 \, \mathrm{H_2O}\)[/tex]. This matches, so the O atoms are balanced:
[tex]\[ \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \, \mathrm{AlCl_3} + 3 \, \mathrm{H_2O} \][/tex]
Thus, the balanced equation is:
[tex]\[ \mathrm{Al_2O_3} + 6 \, \mathrm{HCl} \longrightarrow 2 \mathrm{AlCl_3} + 3 \mathrm{H_2O} \][/tex]
Therefore, the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] are:
[tex]\[ (a, b, c, d) = (1, 6, 2, 3) \][/tex]
This matches with option (A).
So, the correct answer is:
(A) [tex]\(1, 6, 2, 3\)[/tex]
