Is 0.8 mol of NaOH enough if it is known that it interacts with Fe (NO3) 3 weighing 48.3 g.

Given:
n (NaOH) = 0.8 mol
m (Fe (NO3) 3) = 48.3 g

To find:
n (NaOH) – Enough?

Decision:
1) 3NaOH + Fe (NO3) 3 => Fe (OH) 3 + 3NaNO3;
2) M (Fe (NO3) 3) = Mr (Fe (NO3) 3) = Ar (Fe) * N (Fe) + Ar (N) * N (N) + Ar (O) * N (O) = 56 * 1 + 14 * 2 + 16 * 3 * 3 = 228 g / mol;
3) n (Fe (NO3) 3) = m (Fe (NO3) 3) / M (Fe (NO3) 3) = 48.3 / 228 = 0.21 mol;
4) n required (NaOH) = n (Fe (NO3) 3) * 3 = 0.21 * 3 = 0.63 mol;
5) n (NaOH)> n required (NaOH).

Answer: The amount of NaOH substance (0.8 mol) will be enough for the interaction.