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

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )