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 just don't understand how to get the answer. Can someone please show me and explain to me how to do this.
I Think the answer is Na2CO3 but its a 1:1 ratio but I don't know if I need to find the molar masses of each compound? HELP PLEASE!
http://www.jiskha.com/display.cgi?id=1361814261
To determine the mass of precipitate formed in the reaction between Na2CO3 and NiCl2, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between Na2CO3 and NiCl2 is:
Na2CO3 + NiCl2 -> NiCO3 + 2NaCl
Step 2: Calculate the number of moles of each reactant.
Given:
- 0.100 mol Na2CO3
- 0.100 mol NiCl2
Step 3: Determine the stoichiometry of the reaction.
By examining the balanced chemical equation, we see that the molar ratio between Na2CO3 and NiCO3 is 1:1. Therefore, 1 mole of Na2CO3 reacts with 1 mole of NiCO3.
Step 4: Identify the limiting reactant.
To identify the limiting reactant, compare the moles of each reactant to the stoichiometry of the reaction. Since the molar ratio of Na2CO3 to NiCO3 is 1:1, the reactant that produces the smallest amount of product is the limiting reactant.
In this case, both reactants have the same number of moles (0.100 mol). Therefore, Na2CO3 is the limiting reactant because it is present in the smaller amount.
Step 5: Calculate the mass of the precipitate formed.
To calculate the mass of the precipitate (NiCO3) formed, we need the molar mass of NiCO3.
The molar mass of NiCO3 = molar mass of Ni + molar mass of C + 3*(molar mass of O)
= 58.69 g/mol + 12.01 g/mol + 3*(16.00 g/mol)
= 58.69 g/mol + 12.01 g/mol + 48.00 g/mol
= 118.70 g/mol
Since the stoichiometry of the reaction is 1:1 between Na2CO3 and NiCO3, the moles of NiCO3 formed are also 0.100 mol.
To calculate the mass of the precipitate (NiCO3) formed, use the equation:
mass = moles × molar mass
mass of precipitate (NiCO3) = 0.100 mol × 118.70 g/mol
mass of precipitate (NiCO3) = 11.87 g
Therefore, the mass of precipitate formed in the reaction is 11.87 grams.