Đáp án:
\(\begin{array}{l}
3.1)\\
b)\\
{m_{Al}} = 2,7g\\
c)\\
{m_{AlC{l_3}}} = 26,7g\\
d)\\
{m_{CuO}} = 24g\\
3)\\
1)\\
a)\\
{m_{Al}} = 1,7g\\
b)\\
{m_{A{l_2}{O_3}}} = 20,4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
3.1)\\
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{Al}} = \dfrac{m}{M} = \frac{{8,1}}{{27}} = 0,3\,mol\\
{n_{HCl}} = \dfrac{m}{M} = \frac{{21,9}}{{36,5}} = 0,6\,mol\\
\dfrac{{0,3}}{2} > \dfrac{{0,6}}{6} \Rightarrow\text{ Al dư} \\
{n_{Al}} \text{ dư}= 0,3 - \dfrac{{0,6 \times 2}}{6} = 0,1\,mol\\
{m_{Al}} \text{ dư}= n \times M = 0,1 \times 27 = 2,7g\\
c)\\
{n_{AlC{l_3}}} = \dfrac{{0,6 \times 2}}{6} = 0,2\,mol\\
{m_{AlC{l_3}}} = n \times M = 0,2 \times 133,5 = 26,7g\\
d)\\
{n_{{H_2}}} = \dfrac{{0,6 \times 3}}{6} = 0,3\,mol\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = {n_{{H_2}}} = 0,3\,mol\\
{m_{CuO}} = n \times M = 0,3 \times 80 = 24g\\
\text{ Câu }3)\\
1)\\
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{12,5}}{{27}} \approx 0,46\,mol\\
n{O_2} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
\dfrac{{0,46}}{4} > \dfrac{{0,3}}{3} \Rightarrow \text{ Al dư}\\
{n_{Al}} \text{ tham gia}= \dfrac{{0,3 \times 4}}{3} = 0,4\,mol\\
{m_{Al}} \text{ dư}= 12,5 - 0,4 \times 27 = 1,7g\\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{{0,3 \times 2}}{3} = 0,2\,mol\\
{m_{A{l_2}{O_3}}} = n \times M = 0,2 \times 102 = 20,4g
\end{array}\)
