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m AgCl=?; N=2 moli NaCl
μNaCl = 23+35.5 = 58.5
m NaCl = N * μ = 2* 58.5 = 117g/mol
μAgCl= 108+35.5 = 143.5
58.5 143.5
AgNO3 + NaCl => AgCl + NaNO3
117 x
x=117*143.5/58.5 = 287 g
Răspuns:
MNaCl = 23+ 35,5=58,5g/mol
MAgCl=108×35,5=143,5g/mol
m=58,5×2=117g NaCl
58,5g 143,5g
AgNO3 + NaCl => AgCl + NaNO3
117g xg
x=117×143,5/58,5=calculează.