A solution containing 0.100 mol of Na2CO3 and 0.100 mol of NiCl2 is allowed to react. What mass of precipitate forms in this reaction?
What have done so far. I showed you how to convert mols Na2CO3 to mols NiCO3.
You convert mols NiCl2 to mols NiCO3 the same way, then take the smaller of the two mols as the correct one.
So it would be NiCl2 right?
To determine the mass of precipitate formed in the reaction, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
The balanced chemical equation for the reaction between Na2CO3 and NiCl2 is:
2 Na2CO3 + NiCl2 -> 2 NaCl + NiCO3
From the balanced equation, we can see that 2 moles of Na2CO3 react with 1 mole of NiCl2 to produce 1 mole of NiCO3.
Given that we have 0.100 mol of both Na2CO3 and NiCl2, we can calculate the moles of NiCO3 formed from each reactant.
For Na2CO3:
0.100 mol Na2CO3 x (1 mol NiCO3 / 2 mol Na2CO3) = 0.050 mol NiCO3
For NiCl2:
0.100 mol NiCl2 x (1 mol NiCO3 / 1 mol NiCl2) = 0.100 mol NiCO3
From the calculations, we can see that the amount of NiCO3 formed is limited by the amount of Na2CO3 present because Na2CO3 produces less NiCO3. Therefore, Na2CO3 is the limiting reactant.
Now, we can calculate the mass of the precipitate (NiCO3) formed using the molar mass of NiCO3. The molar mass of NiCO3 is the sum of the atomic masses of nickel (Ni), carbon (C), and three oxygen (O) atoms:
Molar mass of NiCO3 = (atomic mass of Ni) + (atomic mass of C) + (3 × atomic mass of O)
Molar mass of NiCO3 = 58.69 g/mol + 12.01 g/mol + (3 × 16.00 g/mol)
= 164.69 g/mol
Now, we can calculate the mass of NiCO3 formed using the equation:
Mass = moles × molar mass
Mass = 0.050 mol × 164.69 g/mol
Mass = 8.235 g
Therefore, the mass of precipitate (NiCO3) formed in this reaction is 8.235 grams.