M MgSO4=120g/mol
MCuSO4= 160g/mol
consideram x g MgSO4 si y g CuSO4 ==> x+y=70
120g MgSO4......32g S
x g .....................a=32x/120 g S
160g CuSO4..........32 g S
y g........................b=32y/160 g S
100 g amestec..............22,85 g S
70g amestec.................[32x/120+32y/160]
faci produsul mezilor=produsul extremilor si simplificari ==>
80x+60y=4800 la care adaugi ecuatia :
x+y=70
rezolvi sistemul si ==> x=30 g y=40 g
70g amestec......30 g MgSO4........40g CuSO4
100 g amestec.......z.........................w ==> z=.....%..; w=......%.
b) n MgSO4=30/120=0,25 moli
n CuSO4=40/160=0,25 moli
nMgSO4/nCuSO4=0,25/0,25
raportul molar este 1:1