For the reaction: Pb(NO3)2(aq) + 2KCl(aq)->PbCl2(s) + 2KNO3(aq)
How many grams of PbCl2 will be formed from 63.0 mL of a 1.60 M KCl solution?
mols KCl = M x L = ?
mols PbCl2 formed = ?mols KCl x (1 mol PbCl2/2 mols KCl) = ?mols KCl x 1/2 = ? mols PbCl2.
Then grams PbCl2 = mols PbCl2 x molar mass PbCl2.
To find out how many grams of PbCl2 will be formed from 63.0 mL of a 1.60 M KCl solution, we need to use stoichiometry and molar ratios.
Here's how you can solve it step-by-step:
1. Look at the balanced chemical equation provided: Pb(NO3)2(aq) + 2KCl(aq) -> PbCl2(s) + 2KNO3(aq)
2. Determine the molar ratio between KCl and PbCl2 from the balanced equation. In this case, the ratio is 2:1. This means that for every 2 moles of KCl, 1 mole of PbCl2 is formed.
3. Convert the volume of the KCl solution to moles. To do this, you need to know the molarity (M) of the solution. In this case, the molarity is given as 1.60 M. The formula to calculate moles is:
Moles (n) = Molarity (M) x Volume (L)
Conversion factor: 1 mL = 0.001 L (volume to liters conversion)
n = 1.60 mol/L x 0.0630 L = 0.1008 moles
Therefore, 63.0 mL of the 1.60 M KCl solution is equal to 0.1008 moles of KCl.
4. Apply the molar ratio from step 2 to calculate the number of moles of PbCl2 formed. Since the ratio is 2:1, and we have 0.1008 moles of KCl, we can calculate the number of moles of PbCl2:
Moles of PbCl2 = (0.1008 moles KCl) x (1 mole PbCl2 / 2 moles KCl) = 0.0504 moles PbCl2
5. Convert the moles of PbCl2 to grams. To do this, you need to know the molar mass of PbCl2, which is the sum of the atomic masses of lead (Pb) and chlorine (Cl):
Molar mass of PbCl2 = (207.2 g/mol) + (2 x 35.45 g/mol) = 278.1 g/mol
Mass (grams) = Moles (n) x Molar mass (M)
Mass of PbCl2 = 0.0504 moles x 278.1 g/mol = 13.99 grams
Therefore, 63.0 mL of a 1.60 M KCl solution will produce approximately 13.99 grams of PbCl2.