8) How many grams of lead (II) chloride can be made from 25.0 mL of 0.500 M potassium chloride solution reacting with excess lead (II) nitrate solution? The molar mass of PbCl2 is 278.1 g/mol

Answer 1

First, you must start with a balanced chemical equation.

2KCl + Pb(NO₃)₂ --> PbCl₂ + 2KNO₃

There is a lot going on in this question. You are expected to know that the roman number II is the charge of the lead cation. Chlorine has a charge or -1 so the proper formula for lead II chloride is PbCl₂. Same for potassium chloride...potassium is a +1 charge and chlorine again is -1; giving the formula KCl. Lead II nitrate; lead is a +2 charge while the nitrate is a polyatomic ion with a -1 charge. To balance the charges in the compound, you must have 2 nitrate ions that bond with one lead ion giving...Pb(NO₃)₂.

You are also expected to know the formula for concentration (M) as: Molarity = moles of solute/Liters of solvent.

In order to now how much KCl to start with you must rearrange the molarity equation to solve for moles: (M)(V). ALSO, before you multiply that, you have to convert the mL into L. Ultimately this would give you (0.5M)(0.025L) = 0.0125moles of KCl.

NOW, set this up in a dimensional analysis problem starting with the 0.0125moles KCl x 1molePbCl₂/2 mole KCL x 278.1g PbCl₂/1 mol PbCl_2. Moles of KCl cancel and moles of PbCl₂ cancel leaving gram of PbCl₂ = 1.74g when you round to the appropriate number if significant digits.

This is a long problem with several different layers of understanding. Gotta love stoichiometry!