If 23 grams of iron (ID chloride reacts with 41 grams of sodium phosphate, what is the limiting reagent? How much sodium chloride can be formed?

Answer 1

One easy way to identify the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation, and whichever value is less represents the limiting reactant. So first, we need a balanced equation. Your question doesn't make sense, so I'll assume the reaction is between iron(II) chloride and sodium phosphate:

3FeCl₂ + 2Na₃PO₄ ==> Fe₃(PO₄)₂ + 6NaCl .. balanced equation

molar mass FeCl₂ = 127 g / mol

molar mass Na₃PO₄ = 164 g / mol

Dividing mols of each by their coefficient:

23 g FeCl₂ x 1 mol / 127 g = 0.181 mols FeCl₂ (÷3->0.06)

41 g Na₃PO₄ x 1 mol / 164 g = 0.250 mols Na₃PO₄ (÷2->0.125)

Since 0.06 is less than 0.125, FeCl2 is the limiting reactant, and the moles of FeCl₂ (0.181 mols) will determine how much NaCl can be formed, as shown below:

mols NaCl formed = 0.181 mol FeCl₂ x 6 mols NaCl / 3 mols FeCl₂ = 0.362 mols NaCl formed

mass NaCl formed = 0.362 mols NaCl x 58.4 g NaCl / mol = 21.1 g NaCl formed