Iron(ii) chloride and sodium carbonate react to make iron(ii) carbonate and sodium chloride: fecl2(aq) + na2co3(s) → feco3(s) + 2nacl(aq). given 1.24 liters of a 2.00 m solution of iron(ii) chloride and unlimited sodium carbonate, how many grams of iron(ii) carbonate can the reaction produce?

287.32 g.

Explanation:

The balanced equation of the reaction is:

FeCl₂ + Na₂CO₃ → FeCO₃ + 2NaCl.

It is clear that 1.0 mole of FeCl₂ reacts with 1.0 mole of Na₂CO₃ to produce 1.0 mole of FeCO₃ and 2.0 moles of NaCl.The limiting reactant is FeCl₂ and Na₂CO₃ is an unlimited reactant.We can calculate the no. of moles of FeCl₂ using the data available:

M = 2.0 mol/L and V = 1.24 L.

∴ n of FeCl₂ = M x V = (2.0 mol/L)(1.24 L) = 2.48 mol.

Using cross multiplication; we can calculate the no. of moles of  iron(II) carbonate can the reaction produce:

1.0 mole of FeCl₂ produces → 1.0 mole of FeCO₃, from the stichiometry.

2.48 moles of FeCl₂ produces → ??? mole of FeCO₃.

The no. of moles of  iron(II) carbonate can the reaction produce = (1.0 mol)(2.48 mol) / (1.0 mol) = 2.48 mol.Now, we can calculate the grams (mass) of iron(II) carbonate can the reaction produce = n x molar mass = (2.48 mol)(115.854 g/mol) = 287.32 g.

How do we find the answer?