If an experiment with 10.2 g barium chloride produced 14.5 g silver chloride, calculate thr experimental mole ratio of silver chloride to barium chloride

The experimental mole ratio of silver chloride to barium chloride is calculated as below

fin the mole of each compound

mole = mass/molar mass

moles of AgCl = 14.5g/142.5 g/mol = 0.102 moles of AgCl

moles of BaCl2 = 10.2 g/208 g/mol = 0.049 moles of BaCl2

find the mole ratio by dividing each mole with the smallest mole (0.049)

AgCl = 0.102/0.049 = 2

BaCl2 = 0.049/0.049 = 1

therefore the mole ratio AgCl to BaCl2 is 2 : 1