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Question
Balance the following equation:
\[\ce{K + H2O->KOH + H2}\]
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Solution
\[\ce{K + H2O->KOH + H2}\]
Multiply H2O by 2 on left side and KOH by 2 on right to make H and O equal.
Atoms of Na, O, and H are balanced on both sides.
\[\ce{K + 2H2O->2KOH + H2}\]
To make K equal make 2K on left.
\[\ce{2K + 2H2O->2KOH + H2}\]
There are 2K atoms, 2(O) atoms and 4H atoms on both side.
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