Donald asks…if 30.0mL of 0.150 M CaCl2 is added to 15.0mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate?a. 0.430g
b. 0.645g
c. 0.215g
d. 2.64gNancy answers:0.0300 L of 0.150 mol/litre CaCl2 = 0.0045 moles CaCL2
0.0150 L of 0.100 mol/litre AgNO3 = 0.0015 moles AgNO3
@ 1 CaCl2 & 2 AgNO3 --> 2 AgCl & 1 Ca(NO3)2
because they need twice as many moles of AgNO3 to react, and did not provide it, AgNO3 is the limiting reagent
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1 mole AgNO3 --> 1
