How many grams of NiCO3 will be formed when a 0.250 M NiCl2 solution reacts completely with 28.3 mL of a 0.129 M Na2CO3 solution?

The answer is 0.433g, but I don't understand how.

0.433 g of NiCO₃ are formed

Explanation:

We determine the reaction:

NiCl₂ + Na₂CO₃ → NiCO₃ + 2NaCl

Ratio is 1:1, 1 mol of chloride reacts with 1 mol of sodium carbonate in order to produce 1 mol of nickel carbonate . Let's determine the moles we used of sodium carbonate.

We convert the volume from mL to L → 28.3 mL . 1L/1000mL = 0.0283mL

Volume . Molarity = Moles → 0.0283L . 0.129mol/L = 3.65×10⁻³ moles

Certainly, the limiting reactant is the sodium carbonate.

I used 0.250 moles of chloride and I need the same amount of carbonate, but I only have 3.65×10⁻³ moles.

As ratio is 1:1, 3.65×10⁻³ moles of sodium carbonate must produce 3.65×10⁻³ moles of nickel carbonate.

We convert the moles to mass →

3.65×10⁻³ moles . 118.69 g / 1 mol = 0.433 g

The answer to your question is 0.433 g

Explanation:

Data

mass of NiCO₃ = ?

concentration of NiCl₂ = 0.25 M

Concentration of  Na₂CO₃ = 0.129 M

Volume of Na₂CO₃ = 0.129 M

Balanced chemical reaction

          NiCl₂  + Na₂CO₃   ⇒   NiCO₃  +  2NaCl

Molecular weight  NiCO₃ = 118.7 g

Molecular weight Na₂CO₃ = 106 g

Process

1.- Calculate the number of moles of Na₂CO₃

Moles = Molarity x Volume

Volume = 28,3 ml = 0.0283 L

Moles = 0.129 x 0.0283

Moles = 0.00365

2.- Convert the moles to grams

                   106 g of Na₂CO₃   1 mol

                      x                           0.00365 moles

                     x = (0.00365 x 106) / 1

                     x = 0.3869 g

3.- Calculate the grams of NiCO₃

                    106 g of Na₂CO₃  118.7 g of NiCO₃

                     0.3869 g              x

                      x = (0.3869 x 118.7)/106

                      x = 0.433 g

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