Balance the following reaction: \(3\,\text{MnO}_2\,(s) + x\,\text{Al}\,(s) \rightarrow 3\,\text{Mn}\,(l) + y\,\text{Al}_2\text{O}_3\,(s) + \text{Heat}\)
x = 4, y = 2
This is a thermite-type displacement reaction, where aluminium reduces a metal oxide to the free metal while itself getting oxidised to aluminium oxide, releasing a large amount of heat. To balance it, the number of atoms of each element must be equal on both sides.
Count the oxygen atoms first. On the left, \(3\,\text{MnO}_2\) has \(3 \times 2 = 6\) oxygen atoms. On the right, each \(\text{Al}_2\text{O}_3\) has 3 oxygen atoms, so \(y\) units give \(3y\) oxygen atoms. Setting \(3y = 6\) gives \(y = 2\).
Now balance aluminium. On the right, \(y = 2\) units of \(\text{Al}_2\text{O}_3\) contain \(2 \times 2 = 4\) aluminium atoms. To match this on the left, \(x = 4\).
Check manganese: 3 Mn on the left and 3 Mn on the right—already balanced. The full balanced equation is \(3\,\text{MnO}_2 + 4\,\text{Al} \rightarrow 3\,\text{Mn} + 2\,\text{Al}_2\text{O}_3 + \text{Heat}\).
Hence, x = 4 and y = 2.
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