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?
I think I know how to do it but don't understand. Do I figure out the mass of each of the compounds? HELP please!!!
This is a limiting reagent problem I know that because amounts for BOTH reactants are given.
I do these the long way but I think that is easier. Write and balance the equation.
Convert 0.l00 mol Na2CO3 to NiCO3(the ppt) assuming you have all of the NiCl2 needed.
Then convert 0.100 mol NiCl2 to NiCO3 assuming you have all of the Na2CO3 needed.
These numbers won't agree; the correct one is ALWAYS the smaller value. That will be the mole NiCO3 formed and the reagent producing that number is the limiting reagent. The other reagent will use whatever is needed and there will be some that is unreacted.
Then mols is converted to grams. g = mols x molar mass.
To determine the mass of the precipitate formed in this reaction, you need to understand the concept of stoichiometry and the mole ratio between the reactants and products.
First, you'll need to write the balanced chemical equation for this reaction. The formula for sodium carbonate is Na2CO3, and for nickel(II) chloride it is NiCl2. The balanced equation for the reaction of Na2CO3 with NiCl2 can be represented as:
Na2CO3 + NiCl2 -> NiCO3 + 2NaCl
According to the balanced equation, one mole of Na2CO3 reacts with one mole of NiCl2 to produce one mole of NiCO3 and two moles of NaCl.
Now, let's calculate the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, compare the number of moles of each reactant present.
In this case, both Na2CO3 and NiCl2 are present in equal amounts (0.100 mol each). Since the stoichiometry of the balanced equation is 1:1, both reactants will be completely consumed. Therefore, there is no excess reactant, and we don't need to do any further calculations to determine the limiting reactant.
Next, we calculate the molar mass of the product, NiCO3. The atomic masses are: Ni = 58.69 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol. The molar mass of NiCO3 is 58.69 + 12.01 + (3 * 16.00) = 118.70 g/mol.
Given that the stoichiometry of the reaction is 1:1, the mass of the precipitate formed (NiCO3) will be equal to the molar mass (118.70 g/mol).
Therefore, the mass of the precipitate formed in this reaction is 118.70 g.