CaCl2 + Na2CO3 ---> CaCO3 +2NaCl
one to one in mols of each
75 *.6 = x * .4
a) What volume of 0.40M calcium chloride solution would be required?
one to one in mols of each
75 *.6 = x * .4
CaCO3 = 100 g/mol
so
4.5 grams
Step 1: Calculate the moles of sodium carbonate.
We know that the concentration of the sodium carbonate solution is 0.60M, which means there is 0.60 moles of sodium carbonate in 1 liter (1000 mL) of the solution. Therefore, we can calculate the moles of sodium carbonate present in 75.0mL using the following formula:
moles = concentration ร volume
moles of sodium carbonate = 0.60M ร (75.0mL / 1000mL)
Step 2: Calculate the moles of calcium chloride.
From the balanced chemical equation, we know that the stoichiometric ratio between sodium carbonate and calcium chloride is 1:1. Thus, the moles of calcium chloride required will be the same as the moles of sodium carbonate obtained in Step 1.
moles of calcium chloride = moles of sodium carbonate
Step 3: Convert moles of calcium chloride to volume of 0.40M calcium chloride solution.
Since we know the concentration of the calcium chloride solution is 0.40M, we can use the following formula to calculate the volume required:
volume = moles / concentration
volume of 0.40M calcium chloride solution = moles of calcium chloride / 0.40M
Substituting the value from Step 2 into the equation, we can determine the volume required.
It's important to note that the final answer will be in liters, so if you want to convert it to a more common unit like milliliters, you can multiply it by 1000.
By following these steps, you can determine the volume of 0.40M calcium chloride solution required to react completely with 75.0mL of 0.60M sodium carbonate solution.