Question

B. Calculate the grams of lead(II) iodide that can be produced from 5.00 moles of potassium iodide.

Answer 1

Answer:-

$2Kl +Pb(NO_3)_2 \rightarrow PbI_2 + 2KNO_3$

1 mole potassium iodide gives 1 mole lead(II) iodide

So 5 mole of potassium iodide will give 5 mole of lead(II) iodide

$Mass = moles ×molar \ mass$

= 5×461.01

$=2305.05$ g answer