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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Answered by kenmyna
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