For the reaction represented by the equation pb(no3)2 + 2ki → pbi2 + 2kno3, how many moles of lead(ii) iodide are produced from 300. g of potassium iodide and an excess of pb(no3)2? a. 11.0 mol selected: b. 1.81 mol c. 3.61 mol d. 0.904 mol

Number of moles equals of potassium iodide;

The molar mass of potassium iodide is 166 g/mole

Moles = 300/166

            = 1.8072moles

According to the equation;

2 moles of KI produces 1 mole of PbI2 9lead (ii) iodide

Therefore; the number of moles of lead (ii) iodide produced will be;

                = 1.8072/2

                = 0.9036moles

Thus the number of moles of lead (ii) iodide is 0.904 mole

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.