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

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