Đáp án:

Câu 10:

\(\begin{array}{l} a,\ V=V_{O_2}=11,2\ (l).\\ b,\ m_{KClO_3\ \text{(thực tế)}}=44,913\ (g).\end{array}\)

Giải thích các bước giải:

Câu 9:

\(\begin{array}{l} PTHH:\\ 4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\\ 4P+5O_2\xrightarrow{t^o} 2P_2O_5\\ CH_4+2O_2\xrightarrow{t^o} CO_2\uparrow+2H_2O\\ S+O_2\xrightarrow{t^o} SO_2\end{array}\)

Câu 10:

\(\begin{array}{l} a,\\ PTHH:2C_2H_2+5O_2\xrightarrow{t^o} 4CO_2\uparrow+2H_2O\\ n_{C_2H_2}=\dfrac{4,48}{22,4}=0,2\ (mol).\\ Theo\ pt:\ n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,5\ (mol).\\ \Rightarrow V=V_{O_2}=0,5\times 22,4=11,2\ (l).\\ b,\\ PTHH:2KClO_3\xrightarrow{t^o} 2KCl+3O_2\uparrow\\ Theo\ pt:\ n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\ (mol).\\ \Rightarrow m_{KClO_3\ \text{(lí thuyết)}}=\dfrac{1}{3}\times 122,5=40,83\ (g).\\ \Rightarrow m_{KClO_3\ \text{(dư)}}=40,83\times 10\%=4,083\ (g).\\ \Rightarrow m_{KClO_3\ \text{(thực tế)}}=40,83+4,083=44,913\ (g).\end{array}\)

chúc bạn học tốt!

Em tham khảo nha :

\(\begin{array}{l}
9)\\
4Al + 3{O_2} \xrightarrow{t^0} A{l_2}{O_3}\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
10)\\
a)\\
2{C_2}{H_2} + 5{O_2} \to 4C{O_2} + 2{H_2}O\\
{n_{{C_2}{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{5}{2}{n_{{C_2}{H_2}}} = 0,5mol\\
{V_{{O_2}}} = n \times 22,4 = 0,5 \times 22,4 = 11,2l\\
b)\\
2KCl{O_3} \to 2KCl + 3{O_2}(1)\\
{n_{KCl{O_3}(1)}} = \dfrac{2}{3}{n_{{O_2}}} = \dfrac{1}{3}mol\\
{m_{KCl{O_3}(1)}} = n \times M = \dfrac{1}{3} \times 122,5 = 40,83g\\
{m_{KCl{O_3}}} = \dfrac{{{m_{KCl{O_3}(1)}} \times 110\% }}{{100\% }} = 44,913g
\end{array}\)