Given: NaCl + AgNO3 -> AgCl + NaNO3
How many moles of AgCl will be produced from 83.0 g of AgNO3, assuming NaCl is available in excess?

The answer to your question is  0.49 moles of AgCl

Explanation:

Data

mass of AgNO₃ = 83 g

moles of AgCl = ?

mass of NaCl = excess

Balanced chemical reaction

              NaCl  +  AgNO₃  ⇒  AgCl  +  NaNO₃

Process

1.- Calculate the molar mass of AgNO₃

AgNO₃ = 108 + 14 + (16 x 3) = 170 g

2.- Convert the grams of AgNO₃ to moles

                   170 g of AgNO₃ 1 mol

                     83 g of AgNO₃  x

                          x = (83 x 1) / 170

                          x = 0.49 moles

3.- Calculate the moles of AgCl

                              1 mol of AgCl  1 mol of AgNO₃

                              x mol of AgCl 0.49 moles of AgNO₃

                                  x = (0.49 x 1)/1

                                  x = 0.49 moles of AgCl

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