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a fost răspuns

Cati moli se gasesc in
190 g MgCl2
492 g Ca(NO3)2
342 g Al2(SO4)3

Răspuns :

a)M(MgCl2) = 24 +(2x35,5)= 95g/mol
v(MgCl2)=m/M=190g/95g/mol= 2mol
b)M(Ca(NO3)2)= 40 +(2x14) +(6x16)=164g/mol
m(Ca(NO3)2) =m/M=492g/164g/mol=3mol
c) M(Al2(SO4)3)=(2 x 27) +(3 x 32) +(12x16)= 342g/mol
m(Al2(SO4)3)=m/M= 342g/342g/mol=1mol

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