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