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

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

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