Calculate what mass of iron (III) hydroxide is formed by the interaction of iron (III) sulfate with 4.8 g of sodium hydroxide.

So, let’s write the reaction equation:
Fe2 (SO4) 3 + 6NaOH = 3 Na2SO4 + 2 Fe (OH) 3
Theoretically for sodium hydroxide:
n = 6 mol (look at the coefficients)
M = 40 g / mol (according to the periodic table, we look at the atomic masses of 3 elements)
m = 240g (40 * 6 = 240)
Theoretically for iron hydroxide 3:
n = 2 mol
M = 107 g / mol
m = 214 g (107 * 2 = 214)
Let’s make the proportion: (by weight practically / theoretically)
4.8 / 240 = x / 214
x = 4.28
m (Fe (OH) 3) = 4.28 g
Answer: 4.28 g.