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? the reaction can produce grams of iron(ii) carbonate.

              287.30 g of FeCO₃

Solution:

The Balance Chemical Equation is as follow,

                            FeCl₂ + Na₂CO₃    →    FeCO₃ + 2 NaCl

Step 1: Calculate Mass of FeCl₂ as,

                             Molarity  =  Moles ÷ Volume

Solving for Moles,

                             Moles  =  Molarity × Volume

Putting Values,

                             Moles  =  2 mol.L⁻¹ × 1.24 L

                            Moles  =  2.48 mol

Also,

                             Moles  =  Mass ÷ M.Mass

Solving for Mass,

                             Mass  =  Moles × M.Mass

Putting Values,

                             Mass  =  2.48 mol × 126.75 g.mol⁻¹

                             Mass =  314.34 g of FeCl₂

Step 2: Calculate Mass of FeCO₃ formed as,

According to equation,

           126.75 g (1 mole) FeCl₂ produces  =  115.85 g (1 mole) FeCO₃

So,

                314.34 g of FeCl₂ will produce  =  X g of FeCO₃

Solving for X,

                      X =  (314.34 g × 115.85 g) ÷ 126.75 g

                      X =  287.30 g of FeCO₃

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.