QuestionDetermine the limiting reagent and maximum moles of silver chloride from 0.581 moles of silver nitrate and 0.195 moles of copper(II) chloride.

Studdy Solution
Calculate the maximum amount of AgCl that can be produced using the amount of the limiting reagent (CuCl2).
$$Amount_{AgCl} = Amount_{CuCl2} =0.195\, moles$$The limiting reagent is CuCl2 and the maximum amount of AgCl that can be produced is0.195 moles.