Question

Calculate the grams of lead (III) iodide that can be produced from 5.00 moles of potassium iodideI picked A but I’m not sure if it’s correct

Answer 1

Explanation:

The question gives us the following reaction:

Pb(NO3)2 + 2 KI ---> PbI2 + 2KNO3

The equation tells us that 2 moles of potassium iodide (KI) produce 1 mole of lead(II) iodide (PbI2).

So:

2 moles KI ---- 1 mole PbI2

5 moles KI ---- x moles

x = 2.5 moles

Now let's transform 2.5 moles of PbI2 into grams:

m = n*MM

MM of PbI2 = 461 g/mol

m = 2.5*461 = 1,152.5 g

Answer: alternative "D" 1.15 x 10^3 g