Bạn tham khảo!

a) Ta có PTHH sau:

$\text{2Al+6HCl$\rightarrow$2AlCl3+3H2}$

b) Có $\text{nAl=$\dfrac{m}{M}$=$\dfrac{8,1}{27}$=0,3 mol}$

Có $\text{nHCl=$\dfrac{m}{M}$=$\dfrac{21,9}{36,5}$=0,6 mol}$

Vì $\dfrac{0,3}{2}$>$\dfrac{0,6}{6}$

$\rightarrow$ $\text{HCl hết, Al dư}$

Có $\text{nAl dư=0,3-0,2=0,1 mol}$

$\rightarrow$ $\text{mAl dư=0,1.27=2,7g}$

c) Có $\text{nAlCl3=nAl phản ứng=0,2 mol}$

$\rightarrow$ $\text{mAlCl3=n.M=0,2.133,5=26,7g}$

d) $\text{CuO+H2$\xrightarrow{t^o}$Cu+H2O}$

Có $\text{nH2=nCuO=$\dfrac{0,6}{2}$=0,3 mol}$

$\rightarrow$ $\text{mCuO=n.M=0,3.80=24g}$

Đáp án:

\(\begin{array}{l}
b)\\
{m_{Al}} = 2,7g\\
c)\\
{m_{AlC{l_3}}} = 26,7g\\
d)\\
{m_{CuO}} = 24g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\,2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{Al}} = \dfrac{{8,1}}{{27}} = 0,3\,mol\\
{n_{HCl}} = \dfrac{{21,9}}{{36,5}} = 0,6\,mol\\
\dfrac{{{n_{Al}}}}{2} > \dfrac{{{n_{HCl}}}}{6} \Rightarrow Al \text{ dư }\\
{n_{Al}} \text{ dư }= 0,3 - 0,6 \times \dfrac{2}{6} = 0,1\,mol\\
{m_{Al}}  \text{ dư }= 0,1 \times 27 = 2,7g\\
c)\\
{n_{AlC{l_3}}} = 0,6 \times \dfrac{2}{6} = 0,2\,mol\\
{m_{AlC{l_3}}} = 0,2 \times 133,5 = 26,7g\\
d)\\
{n_{{H_2}}} = 0,6 \times \dfrac{3}{6} = 0,3\,mol\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{n_{CuO}} = {n_{{H_2}}} = 0,3\,mol\\
{m_{CuO}} = 0,3 \times 80 = 24g
\end{array}\)