500 ML larutan timbal(II),nitrat =0,1 direaksikan dengan 50 ml larutan asam klorida 0,1 m.berapa gram endapan pbcI2 yang terbentuk jika di ketahui mr pbcI2=278?

V Pb(NO3)2 = 500 mL = 0.5 L
M Pb(NO3)2 = 0.1 M
mol Pb(NO3)2 = 0.1 × 0.5 = 0.05 mol
V HCl = 50 mL = 0.05 L
M HCl = 0.1 M
mol HCl = 0.1 × 0.05 = 0.005 mol

Pb(NO3)2 + 2HCl --> PbCl2 + 2HNO3
a. 0.05 0.005
r. 0.0025. 0.005. 0.0025. 0.005
s
mol PbCl2 = 1/2 × 0.005 = 0.0025 mol
berat PbCl2 = mol × Mr
= 0.0025 x 278
= 0.695 gram