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
Answers
If you are not satisfied with the answer or you can’t find one, then try to use the search above or find similar answers below.
Find similar answers
Chemistry, added 2021-09-16 15:32:46
Chemistry, added 2021-09-16 14:32:46
Chemistry, added 2021-09-12 22:21:52
Chemistry, added 2021-09-12 20:31:21
Chemistry, added 2021-09-03 19:30:02
Chemistry, added 2021-09-03 17:53:06
Chemistry, added 2021-09-02 20:53:32
Chemistry, added 2021-09-02 19:18:58
Chemistry, added 2021-08-29 04:54:01
What Is Used In Ficilitated Diffusion To Assist The Transport Of Sugar And Sodium Molecules ...
Chemistry, added 2021-08-29 03:08:37
What Is Used In Ficilitated Diffusion To Assist The Transport Of Sugar And Sodium Molecules ...
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