<span><span>4<span>K</span></span>+<span>O2 = </span><span>2<span>K2<span>O
K2O+H2O=2KOH
KOH+HCl=KCl+H2O
</span></span></span></span>
СaCO3+H2SO4=CaSO4+H2O+CO2
m(H2SO4)=120*0,05=6г
n(H2SO4)=6/98=0,06моль
n(CO2)=0.06моль
V(CO2)=0.06*22.4=1.344л
Прикрепила................................
Дано:
m(HBr) = 120 г
w(HBr) = 10% = 0,1
Найти:
m(MgBr2) - ?
V(H2) - ?
Решение:
Mg + 2HBr = MgBr2 + H2
n(HBr) = (m(HBr) * w(HBr))/M(HBr) = (120*0,1)/81 = 0,148 моль
n(H2) = n(MgBr2) = n(HBr)/2 = 0,148/2 = 0,074 моль
m(MgBr2) = n(MgBr2) * M(MgBr2) = 0,074 * 184 = 13,616 г
V(H2) = n(H2) * M(H2) = 0,074 * 2 = 0,148 л