Question

_na + _o₂ → ___nacl

Explanation:

Step1: Balance the oxygen atoms

The left - hand side has \(O_2\) (2 oxygen atoms). To balance oxygen in \(NaCl\) (which has no oxygen, but we consider the general balancing), we can first balance the chlorine (since \(Cl\) comes from \(O_2\) in the sense of the reaction with \(Na\)). But more accurately, for the reaction \(Na+O_2
ightarrow NaCl\), there is a mistake in the reactant. Assuming it's \(Na + Cl_2
ightarrow NaCl\) (common reaction). For \(Cl_2\) (2 \(Cl\) atoms), we put a 2 in front of \(NaCl\): \(Na+Cl_2
ightarrow 2NaCl\).

Step2: Balance the sodium atoms

Now, there are 2 \(Na\) atoms on the right - hand side (\(2NaCl\)). So we put a 2 in front of \(Na\) on the left - hand side: \(2Na + Cl_2
ightarrow 2NaCl\)

Answer:

2, 1, 2