Calculate the number of grams of AgCl formed when 0.200 L of 0.200 M AgNO3 reacts with an excess of CaCl2. The equation is: 2 AgNO3(aq) + CaCl2(aq) -------> 2 AgCl(s) + Ca(NO3)2(aq)
Number of mol of AgNO3 reacted = M(AgNO3)*V(AgNO3)
= 0.200 M * 0.200 L
= 0.0400 mol
From reaction,
mol of AgCl formed = mol of AgNO3 reacted
= 0.0400 mol
Molar mass of AgCl,
MM = 1*MM(Ag) + 1*MM(Cl)
= 1*107.9 + 1*35.45
= 143.35 g/mol
use:
mass of AgCl,
m = number of mol * molar mass
= 4*10^-2 mol * 1.434*10^2 g/mol
= 5.734 g
Answer: 5.73 g
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