Дано: AlCl3 = 634 г.
Найти: V(Cl)-?
Решение:
Al+3Cl-->AlCl3
---- ------
3 моль 1 моль
n(AlCl3)=m/M=634/133,5=4,7 моль
M(AlCl3)=27+35,5*3=133,5
V(Cl)=n*Vm=14,1*22,4=315,84 л
Вроде так:)
3.2%
Объяснение:
dilute sulfuric acid reacts with zinc to produce hydrogen
zn + h2so4 = znso4 + h2
find the mass of gas released
n (h2) = n (zn) = m (zn) / mr (zn) = 13/65 = 0.2 mol
then the mass of hydrogen m (h2) = n * mr = 0.2 * 2 = 0.4 g
mass of solution at the end of the reaction
m (solution) = m (solution h2so4) + m (zn) - m (h2) = 1000 + 13-0.4 = 1012.6 gr
we put in a mass of zinc sulfate
n (znso4) = n (zn) = 0.2 mol
m (znso4) = n * mr = 0.2 * (65 + 32 + 64) = 32.2 gr
then the mass fraction of sulfate w = m (znso4) * 100% / m (solution) = 32.2 * 100 / 1012.6 = 3.2%
Fe(OH)3<span>↓</span><span> + 3HCl = FeCl</span>3<span> + 3H</span>2<span>O
</span>Fe(OH)3↓<span>+3H(+)+3Cl(-)=Fe(3+)+3Cl(-)+3H2O
</span>Fe(OH)3↓+3H(+)=Fe(3+)+3H2O