Question

If you reacted 25.00 mL of 0.07 g NaCl with an excess of AgNO3, what mass of AgCl should theoretically form?

Answer 1

first you write the equation: AgNO3 + NaCl -> AgCl + NaNO3
now you see that n(NaCl) = n(AgCl)
n(NaCl)=m(NaCl)/M(NaCl)=1,2 *10^-3 mol
m(AgCl)=n(NaCl)*M(AgCl) = 1,2*10^-3 mol * 143,32 g mol-1 = 0,17 g