2Li + 2H2O = 2LiOH + H2
n(Li) = 1.4/7 = 0,2 моль
n(H2O) = 9/18 = 0.5 моль - избыток
n(H2) = 0.1 моль
V(H2) = 0.1*22.4 = 2.24 л
<span><span>Al<span>(OH)</span>2</span> → <span>AlO</span>+<span>H2O
</span></span><span><span>Fe<span>(OH)</span>2</span> → <span>H2O</span>+<span>FeO
</span></span><span><span>2<span>Fe<span>(OH)</span>3</span></span> → <span>Fe2O3</span>+<span>3<span>H2<span>O
</span></span></span></span>
4 электрона так как протонов 4
A)K2CO3, PbS, Fe(NO3)3,
б)PbCl4, MgPO4, Al(NO3)3