Đáp án:
b) 2,24l
c) 8,76%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
nAl = \dfrac{m}{M} = \dfrac{{2,7}}{{27}} = 0,1\,mol\\
mHCl = 100 \times 7,3\% = 7,3g\\
nHCl = \dfrac{{7,3}}{{36,5}} = 0,2\,mol\\
\dfrac{{0,1}}{2} > \dfrac{{0,2}}{6} \Rightarrow \text{ Al dư}\\
n{H_2} = \dfrac{{0,2}}{2} = 0,1\,mol\\
V{H_2} = n \times 22,4 = 0,1 \times 22,4 = 2,24l\\
c)\\
nAl\,pu = \dfrac{{0,2 \times 2}}{6} = \dfrac{1}{{15}}\,mol\\
m{\rm{dd}}spu = \dfrac{1}{{15}} \times 27 + 100 - 0,1 \times 2 = 101,6g\\
C\% AlC{l_3} = \dfrac{{\frac{1}{{15}} \times 133,5}}{{101,6}} \times 100\% = 8,76\%
\end{array}\)
