Question

if 30 ml of 1.50 M cacl2 is added to 15.0 ml of 0.100 M Agno3, what is the mass in grams of agcl precipitate

Answer 1

Concentration*volume = moles
as long as the units cancel out :)

Answer 2

Using this info, can you calculate the moles of each reactant?

Answer 3

mole of Cl- =\[0.03 \times 1.5 \times 2=0.09\]
mole of Ag+ =\[0.015 \times 0.1=0.0015\]
Ag+ is limiting reactant so we have 0.0015 mole AgCl.
\[mole \times molar mass = mass of perticipate\]
\[0.0015 \times 143.31=0.215 gr\]

Answer 4

CaCl2 + 2 AgNO3 → 2 AgCl + Ca(NO3)2

(30 ml) x (1.50 M CaCl2) = 45 mmol CaCl2
(15.0 ml) x (0.100 M AgNO3) = 1.50 mmol AgNO3
1.50 millimoles of AgNO3 would react completely with 1.50 x (1/2) = 0.75 millimoles of CaCl2, but there is much more CaCl2 present than that, so CaCl2 is in excess and AgNO3 is the limiting reactant.

(0.0150 L) x (0.100 mol/L AgNO3) x (2 mol AgCl / 2 mol AgNO3) x
(143.3212 g AgCl/mol) = 0.215 g AgCl