50 ml of.10 M HNO3 is added to 50 ml of .2M CaCl2. Calculate amount of precipitate formed?
zero. No ppt forms with this mixture unless you exceed the solubility of CaCl2 or Ca(NO3)2 and I don't believe that will happen.
To calculate the amount of precipitate formed, we need to determine the limiting reactant in the reaction between HNO3 and CaCl2. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to write the balanced chemical equation for the reaction between HNO3 and CaCl2:
2HNO3 + CaCl2 → Ca(NO3)2 + 2HCl
From the balanced equation, we can see that 2 moles of HNO3 react with 1 mole of CaCl2 to produce 1 mole of Ca(NO3)2 and 2 moles of HCl.
Next, let's calculate the number of moles of HNO3 and CaCl2 that you have:
Volume of HNO3 = 50 mL = 0.05 L
Molarity of HNO3 = 0.10 M
Number of moles of HNO3 = Molarity x Volume = 0.10 mol/L x 0.05 L = 0.005 mol
Volume of CaCl2 = 50 mL = 0.05 L
Molarity of CaCl2 = 0.20 M
Number of moles of CaCl2 = Molarity x Volume = 0.20 mol/L x 0.05 L = 0.010 mol
Now, to determine the limiting reactant, we compare the mole ratios of HNO3 and CaCl2 in the balanced equation. From the equation, we see that the ratio is 2 moles of HNO3 to 1 mole of CaCl2.
Since the mole ratio of HNO3 to CaCl2 is 2:1, it means that for every 2 moles of HNO3, we need 1 mole of CaCl2 to react completely. However, we have only 0.005 moles of HNO3 and 0.010 moles of CaCl2. This means that there is an excess of CaCl2 and HNO3 is the limiting reactant.
Since 2 moles of HNO3 react with 1 mole of CaCl2, we can conclude that 0.005 moles of HNO3 will react with 0.0025 moles of CaCl2. The molar mass of Ca(NO3)2 is 164 g/mol, so the mass of Ca(NO3)2 precipitate formed can be calculated as follows:
Mass of Ca(NO3)2 = Moles of Ca(NO3)2 x Molar mass of Ca(NO3)2
= 0.0025 mol x 164 g/mol
= 0.41 g
Therefore, the amount of precipitate formed is 0.41 grams.