There are 24.5 moles of silver nitrate used in the following reaction. How many moles of barium nitrate was produced? Equation: 2AgNO3 + BaCl2 → 2AgCl + Ba(NO3) 2

Answer 1

We can see from the reaction stoichiometry that 2 moles of AgNO3 will produce 1 mole of Ba(NO3)2 (provided that silver nitrate is the limiting reactant). So, we can calculate the amount of barium nitrate produced:

moles of Ba(NO3)2 produced = (24.5 moles AgNO3) * [(1 mole Ba(NO3)2) / (2 moles AgNO3)]