posted by Deepa on .
What mass of silver chloride, AgCl will precipitate from a solution that contains 1.50 g of, calcium chloride, CaCl if an excess amount of silver nitrate is present?
CaCl2 + 2AgNO3 --> 2AgCl+ Ca(NO3)2
I keep getting 5!but its wrong...right?
CaCl2 is the limiting reagent. 1 mole of CaCl2 =2 moles of AgNO3=2 moles of AgCl =1 mole of Ca(NO3)2.
1.50g of CaCl2/110.98 g/mol of CaCl2= number of moles of CaCl2
number of moles of CaCl2*(2 moles of AgCl/1 mole of CaCl2) * 143.32 g of AgCl*mole-1= mass of silver chloride.
Didn't punch in the number, but this should be right.