For the chemical reaction 2 ki + pb ( no 3 ) 2 ⟶ pbi 2 + 2 kno3 what mass of lead(ii) iodide is produced from 2.89 mol of potassium iodide

The answer to your question is 1.45 moles of PbI₂

Explanation:

Reaction

                    2KI + Pb(NO₃)₂    ⇒   PbI₂  +  2 KNO₃

Process

1.- Check if the reaction is balanced

                 Reactants           Elements        Products

                         2                  Potassium             2

                         2                  Iodine                    2

                         1                   Lead                       1

                         2                  Nitrogen                2

                        6                   Oxygen                 6

This reaction is balanced

2.- Calculate the moles of Lead (II) iodine using proportions and the coefficients of the balanced reaction.

                         2 mol of KI    1 mol of PbI₂

                       2.89 mol of KI      x

                         x = (2.89 x 1) / 2

                         x = 1.45 moles of PbI₂

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