Giải thích các bước giải:

 a. \(2Al+6HCl\to 2AlCl_3+3H_2\)

b. \(n_{Al}=\dfrac{10,8}{27}=0,4\ (mol)\)

Theo PTHH: \(n_{AlCl_3}=n_{Al}=0,4\ (mol)\)

\(\to m_{AlCl_3}=0,4\cdot 133,5=53,4\ (gam)\)

$n_{H_2}=\dfrac 32\cdot n_{Al}=0,6\ (mol)$

$\to V_{H_2}=0,6\cdot 22,4=13,44$ (lít)

c. Theo PTHH: \(n_{HCl}=3n_{Al}=1,2\ (mol)\)

$\to V_{dd\ HCl}=\dfrac{1,2}{0,5}=2,4$ (lít)

d. \(C_{M\ dd}=\dfrac{0,4}{2,4}=0,167\) (mol/L)

Đáp án:

\(\begin{array}{l} b,\ m_{AlCl_3}=53,4\ g.\\ V_{H_2}=13,44\ lít.\\ c,\ V_{\text{dd HCl}}=2,4\ lít.\\ d,\ C_{M_{\text{dd spư}}}=C_{M_{AlCl_3}}=\dfrac{1}{6}\ M.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ PTHH:\\ 2Al+6HCl\to 2AlCl_3+3H_2\uparrow\\ b,\\ n_{Al}=\dfrac{10,8}{27}=0,4\ mol.\\ Theo\ pt:\ n_{AlCl_3}=n_{Al}=0,4\ mol.\\ \Rightarrow m_{AlCl_3}=0,4\times 133,5=53,4\ g.\\ Theo\ pt:\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,6\ mol.\\ \Rightarrow V_{H_2}=0,6\times 22,4=13,44\ lít.\\ c,\\ Theo\ pt:\ n_{HCl}=3n_{Al}=1,2\ mol.\\ \Rightarrow V_{\text{dd HCl}}=\dfrac{1,2}{0,5}=2,4\ lít.\\ d,\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=2,4\ lít.\\ \Rightarrow C_{M_{\text{dd spư}}}=C_{M_{AlCl_3}}=\dfrac{0,4}{2,4}=\dfrac{1}{6}\ M.\end{array}\)

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