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

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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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