Mg + 2 HCl \(\rightarrow\)MgCl2 + H2
nMg=\(\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo pTHH ta có:
nMg=nH2=0,2(mol)
VH2=22,4.0,2=4,48(lít)
Theo pTHH ta có:
nHCl=2nMg=0,4(mol)
mHCl=36,5.0,4=14,6(g)
mdd HCl 7,3%=\(14,6:\dfrac{7,3}{100}=200\left(g\right)\)
d;Theo PTHH ta có:
nMg=nMgCl2=0,2(mol)
mMgCl2=0,2.95=19(g)
C% dd MgCl2=\(\dfrac{19}{200+4,8-0,2.2}.100\%=9,29\%\)