If 37.5 ml of .100m calcium chloride reacts completely with aqueous silver nitrate, what is the mass of agcl precipitate

The equation is given as follows  
2AgNO3 + CaCl2 yields 2AgCl + Ca(NO3)2 
Since the molarity, M, is equal to the number of moles of solute, n, divided by the volume of the solution, V. It follows from the equation that the number of moles of Cacl2 = 37.5 * .100 * 10^(-3) = 0.00375 moles.  
1 mole of Cacl2 yields 2 moles of Agcl2 
0.00375 moles of Cacl2 will yield X.  
So X = 0.00375 * 2/1 = 0.0075 moles.  
No of moles of Agcl2 = mass /molar mass.  
Molar mass of Agcl = 178.772 
So mass = 178.772 * 0.0075 = 1.34079

The answer would be