Đáp án:

\(  {m_{A{l_2}{O_3}}} = 20,4{\text{ gam}}\)

\( {V_{{O_2}}} = 6,72{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)

Ta có:

\({n_{Al}} = \frac{{10,8}}{{27}} = 0,4{\text{ mol}}\)

\( \to {n_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al}} = 0,2{\text{ mol}}\)

\( \to {m_{A{l_2}{O_3}}} = 0,2.(27.2 + 16.3) = 20,4{\text{ gam}}\)

\({n_{{O_2}}} = \frac{3}{4}{n_{Al}} = 0,3{\text{ mol}}\)

\( \to {V_{{O_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)

\({n_{{H_2}}} = \frac{{15,6}}{{22,4}} = 0,7{\text{ mol}}\)

Phản ứng xảy ra:

\(2{H_2} + {O_2}\xrightarrow{{{t^o}}}2{H_2}O\)

\( \to {n_{{H_2}}} > 2{n_{{O_2}}}\) do vậy \(H_2\) dư

\( \to {n_{{H_2}O}} = 2{n_{{O_2}}} = 0,3.2 = 0,6{\text{ mol}}\)

\( \to {m_{{H_2}O}} = 0,6.18 = 10,8{\text{ gam}}\)

a)4Al+3O2->2Al2O3

b)nAl=10,8/27=0,4(mol)

theo pthh thì:

nAl2O3=1/2nAl=0,4.1/2=0,2(mol)

mAl2O3=0,2.102=20,4(g)

c)theo pthh thì:

nO2=3/4nAl=0,4.3/4=0,3(mol)

vO2=0,3.22,4=6,72(l)

d)nH2O=15,6/22,4=0,696(mol)

2H2+O2->2H2O

theo pthh ta tính nH2O=nO2 (vì H2O dư)

nH2O=nO2=0,3(mol)

mH2O=0,3.18=5,4(g)