mass of silver chloride is obtained by adding excess silver nitrate
solution to a 10.0mL volume of 0.10mol per litre CaCl2 solution?"
2 AgNO3 + CaCl2 -> 2AgCl + Ca(NO3)2
0.01L CaCl2 *(.10 mol CaCl2/ 1 L CaCl2) * (2 mol AgCl/ 1 mol CaCl2) = 0.002 mol of AgCl
we will take the mol of agcl and multiply by themolar mass of Agcl to give you the mass of Agcl obtained
0.002 mol AgCl * (143.32 g AgCl/ 1 mol AgCl) = 0.287 g AgCl
The answer is 0.287gAgcl
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