Question

balance the following equation: (make sure each blank has a number)
(square mathrm{al}_{2} mathrm{o}_{3}+square mathrm{nacl}
ightarrow square mathrm{alcl}_{3}+square mathrm{na}_{2} mathrm{o})

Explanation:

Step1: Count atoms

Count the number of each atom on both sides.
For \(Al\): \(2\) on left (\(Al_2O_3\)), \(1\) on right (\(AlCl_3\)).
For \(O\): \(3\) on left (\(Al_2O_3\)), \(1\) on right (\(Na_2O\)).
For \(Na\): \(1\) on left (\(NaCl\)), \(2\) on right (\(Na_2O\)).
For \(Cl\): \(1\) on left (\(NaCl\)), \(3\) on right (\(AlCl_3\)).

Step2: Balance \(Al\)

Let's start with \(Al\). To balance \(Al\), put a \(2\) in front of \(AlCl_3\).
\(Al_2O_3+NaCl
ightarrow2AlCl_3 + Na_2O\)

Step3: Balance \(Cl\)

Now, for \(Cl\). Since there are \(6\) \(Cl\) atoms on the right (\(2\times3\) in \(2AlCl_3\)), put a \(6\) in front of \(NaCl\).
\(Al_2O_3 + 6NaCl
ightarrow2AlCl_3+Na_2O\)

Step4: Balance \(Na\)

For \(Na\), there are \(6\) \(Na\) atoms on the left (\(6\times1\) in \(6NaCl\)). So, put a \(3\) in front of \(Na_2O\) (since \(3\times2 = 6\)).
\(Al_2O_3+6NaCl
ightarrow2AlCl_3 + 3Na_2O\)

Step5: Check \(O\)

On the left, there are \(3\) \(O\) atoms (\(Al_2O_3\)). On the right, in \(3Na_2O\), there are \(3\times1=3\) \(O\) atoms.

Answer:

\(1\) \(Al_2O_3+6\) \(NaCl
ightarrow2\) \(AlCl_3 + 3\) \(Na_2O\)