In: Chemistry
If 30.0 mL of 0.150 M CaCl2 is added to 35.0 mL of 0.100 M AgNO3, what is the mass of the AgCl precipitate?
moles of CaCl2 = 30 x 0.150 / 1000 = 4.5 x 10^-3
moles of AgNO3 = 35 x 0.1 / 1000 = 3.5 x 10^-3
CaCl2 + 2 AgNO3 ------------------> 2 AgCl + Ca(NO3)2
1 2 2
4.5 x 10^-3 3.5 x 10^-3 ?
here limiting reagent AgNO3 . so AgCl formed based on that.
2 mol of AgNO3 --------------- 2 mol of AgCl
3.5 x 10^-3 AgNO3 -------------> ??
moles of AgCl = 3.5 x 10^-3 x 2 / 2 = 3.5 x 10^-3
mass of AgCl = moles x molar mass
= 3.5 x 10^-3 x 143.3
= 0.5 g
mass of AgCl = 0.5 g