If you evaporated 250. mL of a 3.0 M solution of iron (II) nitrite, what mass of iron (II) nitrite would you recover?

Answer 1

Hey Blue! Since we have a volume and concentration of Fe(NO2)2, we can find the mols of Fe(NO2)2:

250 mL * 1 L/1000 mL * 3.0 mol Fe(NO2)2/L = 0.75 mols Fe(NO2)2

Since the question asks for the mass recovered, we can convert the mols to grams by multiplying by the molar mass:

0.75 mols Fe(NO2)2 * 147.86 g Fe(NO2)2/ mol = 110.89 g Fe(NO2)2

Looking at sig figs, since the concentration contains only 2 sig figs, we can truncate (get rid of) the decimal points to get our answer to 2 sig figs, which is 110 g Fe(NO2)2. I hope this helps!