$$2NaHCO_{3(s)} \longrightarrow Na_2CO_{3(s)} + H_2O_{(g)} + CO_{2(g)}$$
From the equation, we can see that for every 2 moles of NaHCO3 decomposed, 1 mole of water is produced.
First, we need to convert the given mass of NaHCO3 to moles:
$$moles \ of \ NaHCO_3 = 2.10 \ g \ / \ 84 g/mol = 0.025 \ mol$$
Now, we can use the mole ratio from the balanced equation to determine the moles of water produced:
$$moles \ of \ H_2O = moles \ of \ NaHCO_3 \times \frac{1 \ mol \ H_2O}{ 2\ mol \ NaHCO_3}= \frac{(0.025 \ mol) (1 \ mol \ H_2O)}{ (2 mol \ NaHCO_3)}=0.0125 \ mol \ H_2O $$
Finally, we can convert the moles of water back to grams:
$$grams \ of \ H_2O = moles \ of \ H_2O \times molar \ mass \ of \ H_2O$$
$$grams \ of \ H_2O=(0.0125 mol)(18 \ g/mol) = 0.225 \ g$$
Therefore, the decomposition of 2.10 g of NaHCO3 produces 0.225 g of water.
