For the reaction represented by the equation Pb(NO3)2 + 2KI → PbI2 + 2KNO3, how many moles of lead iodide are produced from 3.0 moles of potassium iodide when Pb(NO3)2 is in excess?

Respuesta :

Answer:

1.5 moles of PbI2 will be produced

Explanation:

From the balanced chemical equation

2 moles of KI yields 1mole of PBI2

Then 3 moles of KI will yield 1.5 moles of PbI2.....

Answer:

1.5mole

Explanation:

The equation for the reaction is given below:

Pb(NO3)2 + 2KI → PbI2 + 2KNO3

From the equation,

2moles of KI produced 1mole of PbI2.

Therefore, 3moles of KI will produce = 3/2 = 1.5mole of PbI2.

Therefore, 1.5mole of Pbl2 are produced