Answer to Question #159534 in General Chemistry for Ruby Terrence

Question

How many grams of each reactant is needed to produce 50 grams of Fe2(SO4)3

Answer 1

50 grams of Fe₂(SO4)₃ = $\dfrac{\textsf{mass}}{\textsf{molar mass}} = \dfrac{50}{400} = 0.125 \textsf{ moles}$

$Fe_2(SO_4)_3 = 2Fe + 3S + 3O_4$

This means that one mole of Fe₂(SO4)₃ has two moles of Iron atom, three moles of Sulphur atom, and 12 moles of Oxygen atom.

Therefore, 0.125 moles of Fe₂(SO4)₃ has 0.25 moles of Iron atom, 0.375 moles of Sulphur atom, and 1.5 moles of Oxygen atom.

mass of iron atoms in the compound = 0.25 × molar mass = 0.25 × 55.85 = 13.96g

mass of sulfur atoms in the compound = 0.375 × molar mass = 0.375 × 32 = 12g

mass of oxygen atoms in the compound = 1.5 × molar mass = 1.5 × 16 = 24g