Iron weighing 14 g reacted with chlorine. Find the mass of iron (III) chloride if the practical yield is 90%.

Given:
m (Fe) = 14 g;
interaction with Cl2;
w (FeCl3 yield) = 90%
Find: m (FeCl3) -?
Decision:
1) Let’s write the reaction equation:
2Fe + 3Cl2 = 2FeCl3;
In the reaction, the ratio of moles Fe: FeCl3 = 2: 2;
2) Find the amount of substance v (Fe), v (Fe) = 14 g / 56 g / mol = 0.25 mol.
3) Since Fe: FeCl3 = 2: 2, then v (Fe) = v (FeCl3) = 0.25 mol;
4) m (practical) = 0.25 mol * (56 + 35.5 * 3) = 40.625 g.
5) w = m (t) / m (n) * 100%;
m (t) = m (n) * w / 100%;
m (t) = (40.625 g * 90%) / 100% = 36.5625 g.
Answer: m (FeCl3) = 36.5625 g.