Koliko dm3 gasa se oslobodi pri normalnim uslovima u reakciji Mg sa 200cm3, 30% HCl ako je gustina rastvora 1.24g/cm3?
Mg+2HCl------------->MgCl2+H2
mr=ro*V=1.24*200=248g
m(HCl)=w(HCl)*mr=0.30*248=74.4g
n(HCl)=m(HCl)/Mr=74.4/36.5=2.038mol
n(HCl):n(H2)=2:1
n(H2)=n(HCl)*1/2=2.038*1/2=1.019mol
1mol-------------22.4dm3
1.019mol-----------xdm3
x=1.019*22.4/1=22.83dm3 H2
powered by Drupal
Zapremina
Mg+2HCl------------->MgCl2+H2
mr=ro*V=1.24*200=248g
m(HCl)=w(HCl)*mr=0.30*248=74.4g
n(HCl)=m(HCl)/Mr=74.4/36.5=2.038mol
n(HCl):n(H2)=2:1
n(H2)=n(HCl)*1/2=2.038*1/2=1.019mol
1mol-------------22.4dm3
1.019mol-----------xdm3
x=1.019*22.4/1=22.83dm3 H2