B. Calculate the grams of lead(II) iodide that can be produced from 5.00 moles of potassium iodide.
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 \u00d7molar \\ mass"
= 5×461.01
"=2305.05" g answer
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