M(Al+Zn+Mg)=1.28g+HCl v(H2)=0.896 l+NaOH v(H2)=0.56 - вопрос №16741579128208962056 от хорошист540 09.05.2020 00:14

Question

Химия: M(Al+Zn+Mg)=1.28g+HCl v(H2)=0.896 l+NaOH v(H2)=0.56 l w(Al)?w(Zn)?w(Mg)?​

хорошист540 · Answer

2Al + 6HCl → 2AlCl3 + 3H2m(Al) = 270гn(Al)=m/M=270/27=10 моль2 моль Al = 10 моль3 моль H2 = xX=10*3/2=15 моль=n(H2)V(H2)=Va*n=22,4*15=336 лответ : 336 литров....

M(Al+Zn+Mg)=1.28g+HCl v(H2)=0.896 l+NaOH v(H2)=0.56 l w(Al)?w(Zn)?w(Mg)?​

M(Al+Zn+Mg)=1.28g+HCl v(H2)=0.896 l+NaOH v(H2)=0.56 l w(Al)?w(Zn)?w(Mg)?