Question

10,6 g amestec de clorura de magneziu si clorura de sodiu in raport molar 1:2 reactioneaza cu o solutie de azotat de argint 10%.Calculeaza masa de precipitat obtinuta si masa solutiei de azotat de argint utilizata.

Ajutati-ma va rog urgent. La primu rasp dau coroana

Answer 1

MMgCl2=24+35.5*2=95g/mol
MNaCl=23+35.5=58.5g/mol
MAgNO3=108+14+48=170g/mol
MMg(NO3)2=24+2(14+48)=148g/mol
MNaNO3=23+14+48=85g/mol

95x + 58.5y =10.6g ==>95x + 58.5*2x=10.6==>212x=10.6==>x=0.05 moli

x/y=1/2==>y=2x==>y=2*0.05=0.1 moli 

1mol.........2mol.............1mol
MgCl2 + 2AgNo3 --> Mg(NO3)2 + 2AgCl
0.05mol....x=0.1mol.....y=0.05 mol

n=m/M==>m=n*M=0.05*148=7.4g Mg(NO3)2 precipitat

1mol......1mol..........1mol
NaCl + AgNO3 --> NaNO3 + AgCl
0.1mol....0.1mol......0.1mol

n=m/M==>m=n*M=0.1*85=8.5g NaNO3

mprecipitati(NaNO3+Mg(NO3)2)=8.5+7.4=15.9g

nr total de moli AgNO3(nt)=0.1+0.1=0.2 moli AgNO3

n=m/M==>m=n*M=0.2*170=34g AgNO3

Cp=md/ms*100==>ms=md*100/Cp=34*100/10=340g AgNo3