balance the chemical equation below using the smallest possible whole number stoichiometric coefficients…

Explanation:

Step1: Balance Fe atoms

Initially, there is 1 Fe on the left - hand side and 1 Fe on the right - hand side, so Fe is already balanced for now. $Fe(s)+O_2(g)+H_2O(l)\rightarrow Fe(OH)_2(aq)$

Step2: Balance O atoms

There are 2 O in $O_2$ and 2 O in $H_2O$ on the left - hand side and 2 O in $Fe(OH)_2$ on the right - hand side. Let's start by balancing the oxygen atoms considering the overall reaction. The oxygen in $O_2$ and $H_2O$ needs to be distributed to form $Fe(OH)_2$. We know that in $Fe(OH)_2$, for every Fe atom, there are 2 OH groups. We need to adjust the coefficients to balance the oxygen and hydrogen. Let's start with the oxygen in $O_2$. $2Fe(s)+O_2(g)+2H_2O(l)\rightarrow 2Fe(OH)_2(aq)$ Now, check the hydrogen atoms. On the left - hand side, we have $2\times2 = 4$ H atoms from $2H_2O$, and on the right - hand side, we have $2\times2=4$ H atoms in $2Fe(OH)_2$. Also, the Fe atoms are balanced (2 on both sides) and the O atoms are balanced (4 on both sides).

Answer:

$2Fe(s)+O_2(g)+2H_2O(l)\rightarrow 2Fe(OH)_2(aq)$