Đáp án:
Bạn tham khảo nha !
Giải thích các bước giải:
\(\begin{array}{l}
10.\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = 0,2mol\\
a)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,4mol\\
\to {m_{HCl}} = 14,6g\\
b)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = 0,13mol\\
\to {m_{Fe}} = 7,28g
\end{array}\)
\(\begin{array}{l}
11.\\
a)2KCl{O_3} \to 2KCl + 3{O_2}\\
b)\\
{n_{KCl{O_3}}} = 0,6mol\\
\to {n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 0,9mol\\
\to {V_{{O_2}}} = 20,16l
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{KCl}} = {n_{KCl{O_3}}} = 0,6mol\\
\to {m_{KCl}} = 44,7g\\
d)\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
{n_P} = \dfrac{4}{5}{n_{{O_2}}} = 0,72mol\\
\to {m_P} = 22,32g
\end{array}\)
\(\begin{array}{l}
12.\\
Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,25mol\\
a)\\
{n_{Na}} = 2{n_{{H_2}}} = 0,5mol\\
\to {m_{Na}} = 11,5g\\
b){V_{{H_2}}} = 5,6l
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{NaOH}} = 2{n_{{H_2}}} = 0,5mol\\
\to {m_{NaOH}} = 20g\\
d)\\
PbO + {H_2} \to Pb + {H_2}O\\
{n_{Pb}} = {n_{{H_2}}} = 0,25mol\\
\to {m_{Pb}} = 51,75g
\end{array}\)
