Question

in 200 g solutie HCl de concentratie 36.5% se introduce magneziu .Ce cantitati de clorura de magneziu si hidrogen se obtin

Answer 1

mHCl=200g
cHCl=36,5%

2HCl+Mg=MgCl2 + H2

x= masa MgCl2
y=masa H2

c=md/ms * 100 <=> 36,5=md/200 * 100<=> md=36,5 *2=73 g HCl pur.
Dupa care,din reactie rezulta ca x=73*95 totul supra 2*36.5=95g MgCl2 pur.
 Iar y il afli in acelasi mod. y=masa HCl pur * M hidrogen totul supra M HCl.
sper ca te-am ajutat. :* 

Answer 2

Cp=md/ms*100==>md=73g HCl

n=m/M=73/36.5=2mol HCl
1mol..2mol..1mol....1mol
Mg+2HCl->MgCl2 + H2
1mol..2mol...1mol....1mol

n=m/M==>m=n*M=1*2=2g H2

mMgCl2=95g

MHCl=36.5g/mol
MMgCl2=95g/mol
MH2=2g/mol