Suppose you wanted to add just enough 0.40M calcium chloride solution to 75.0mL of 0.60M of sodium carbonate solution for all of the sodium carbonate and calcium chloride to react.
B) what mass of calcium carbonate would be produced (assuming a 100% yield)
4.5 grams, look below at earlier question
To find the mass of calcium carbonate produced, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in a reaction, determining the maximum amount of product that can be formed.
First, let's write the balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and calcium chloride (CaCl2):
Na2CO3 + CaCl2 → CaCO3 + 2NaCl
According to the equation, one mole of sodium carbonate reacts with one mole of calcium chloride to produce one mole of calcium carbonate and 2 moles of sodium chloride.
Next, we need to calculate the number of moles of each reactant present:
For sodium carbonate:
moles = concentration (M) x volume (L)
moles Na2CO3 = 0.60 M x 0.075 L = 0.045 moles
For calcium chloride:
moles CaCl2 = 0.40 M x 0.075 L = 0.030 moles
Since the stoichiometric ratio in the balanced equation is 1:1 for the reactants, we can see that calcium chloride is the limiting reactant since it has fewer moles compared to sodium carbonate.
Now, let's calculate the moles of calcium carbonate formed:
moles CaCO3 = moles CaCl2 = 0.030 moles
Finally, we can calculate the mass of calcium carbonate produced using its molar mass:
mass CaCO3 = moles CaCO3 x molar mass CaCO3
The molar mass of calcium carbonate (CaCO3) is:
40.08 g/mol (mass of Ca) + 12.01 g/mol (mass of C) + (3 x 16.00 g/mol) (mass of O) = 100.09 g/mol
mass CaCO3 = 0.030 moles x 100.09 g/mol ≈ 3.00 g
Therefore, approximately 3.00 grams of calcium carbonate will be produced assuming a 100% yield.