KCl+AgNO3=AgCl+KNO3
Find the amount of substance in moles
v=m/M
v(KCl)=2.235/(35.453+39.098)=2.235/74.551=0.029 mol
v(AgNO3)=7.65/(107.868+15.999*3+14.0067)=7.65/169.8717=0.045 mol
The reaction equation shows that n (AgNO3): n (KCl) = 1: 1
then KCl is in shortage, and AgNO3 is in excess. We further calculate KCl:
v(AgCl)=0.029
m=v*M=0.029*(108+35.5)=0.029*143.5=4.162 g
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