Answer:
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
Answer:
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
