Lead ions can be precipitated from solution with kcl according to the following reaction:
pb2+(aq) + 2kcl(aq) --> pbcl2(s) + 2k+(aq)

when 28.8g kcl is added to a solution containing 25.6g pb2+, a pbcl2 precipitate forms. the precipitate is filtered and dried and found to have a mass of 29.6g.

determine the theoretical yield of pbcl2.

m=

determine the percent yield for the reaction.

34.48 grams = Theoretical yield

The % yield is 85.85 %

Explanation:

Step 1: The balanced equation

Pb2+(aq) + 2KCl(aq) → PbCl2(s) + 2K+(aq)

For 1 mole Pb2+ consumed, we need 2 moles of KCl to produce 1 mole of PbCl2 and 2 moles of K+

Step 2: Given data

Mass of KCl = 28.8 grams

Mass of Pb2 = 25.6 grams

Mass of dried PbCl2 = 29.6 grams

Molar mass of Pb2+ = 207.2 g/mol

Molar mass of KCl = 74.55 g/mol

Molar mass of PbCl2 = 278.1 g/mol

Step 3: Calculate moles of Pb2+

Moles = mass / Molar mass

Moles Pb2+ = 25.6 grams / 207.2 g/mol = 0.124 moles

Step 4: Calculate moles of KCl

Moles = 28.8 grams / 74.55 g/mol = 0.386 moles

Step 5: Calculate limiting reactant

For 1 mole Pb2+ consumed, we need 2 moles of KCl

Pb2+ is the limiting reactant. It will completely ( 0.124 moles) be consumed. There will remain 0 moles

There will react 2* 0.124 = 0.248 moles of KCl. There will remain 0.386 - 0.248 = 0.138 moles of KCl

Step 6: Calculate number of moles of PbCl2

For 1 mole Pb2+ consumed, we need 2 moles of KCl to produce 1 mole of PbCl2 and 2 moles of K+

For 0.124 moles of Pb2+ consumed ,there will be produced 0.124 moles of PbCl2

Step 7: Calculate mass of PbCl2

Mass = Numbers of moles * Molar mass

mass PbCl2 =0.124 moles * 278.1 g/mol = 34.48 grams = Theoretical yield

Step 8: Calculate % yield

% yield = (actual yield / theoretical yield) * 100 = (29.6 grams / 34.48 grams) *100 % = 85.85 %

The % yield is 85.85 %

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

the answer is definitely c.