Question

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.

Answer 1

Answer:

              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

Answer 2

Answer: 287

Explanation: It says down to 3 sig figs. I just took the Plato Test