Question

Determine the mass of lead (II) iodide that will be produced when 139 ml of 4.4M potassium iodide completely reacts with excess lead(II) nitrate

Answer 1

So,

We're given that we have a solution of 139 ml of 4.4M potassium iodide. Before analyzing the reaction, what we could do is to find the moles of potassium iodide using the information above:

Now, we're given that these moles react according to the following reaction:

Notice that the reaction is already balanced.

Now, what we need to do is to use the stoichiometry (coefficients of the reaction):

Those are the moles of PbI2 produced. To find the mass in grams, we multiply by the molar mass of PbI2:

The answer is 140.98g of PbI2.