Giúp mình với đang cần gấp hóa học 9

Đáp án:

\(4.\)

a)

\(CH_4^{} + 2O_2^{}\xrightarrow{{{t^o}}}CO_2^{} + 2H_2^{}O\)

\(C_2^{}H_2^{} + \frac{5}{2}O_2^{}\xrightarrow{{{t^o}}}2CO_2^{} + H_2^{}O\)

\(Ca(OH)_2^{} + CO_2^{} \to CaC{\text{O}}_3^{} + H_2^{}O\)

b)

\(n_{hh}^{} = \frac{{1,68}}{{22,4}} = 0,075\,mol\)

\(n_{CaC{\text{O}}_3^{}}^{} = \frac{{10}}{{100}} = 0,1\,mol\)

\(CH_4^{};x\,mol;\,C_2^{}H_2^{}:y\,mol\)

\(CH_4^{} + 2O_2^{}\xrightarrow{{{t^o}}}CO_2^{} + 2H_2^{}O\)

\(\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\)

\(C_2^{}H_2^{} + \frac{5}{2}O_2^{}\xrightarrow{{{t^o}}}2CO_2^{} + H_2^{}O\)

\(\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2y\)

\(Ca(OH)_2^{} + CO_2^{} \to CaC{\text{O}}_3^{} + H_2^{}O\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\, \leftarrow \,\,\,\,0,1\)

\(n_{CO_2^{}}^{} = x + 2y = 0,1\,mol\)

\(n_{hh}^{} = x + y = 0,075\,mol\)

\( \to {\text{x}} = 0,05\,mol;y\, = 0,025\,mol\)

\(b)V_{CH_4^{}}^{} = \frac{{0,05}}{{0,075}}.100 = 66,67\% \)

\(V_{C_2^{}H_2^{}}^{} = 100 - 66,67 = 33,34\% \)

\(5.\)

\(n_{hh}^{} = \frac{{0,224}}{{22,4}} = 0,001\,mol\)

\(n_{O_2^{}}^{} = \frac{{0,504}}{{22,4}} = 0,0225\)

\(CH_4^{};x\,mol;\,C_2^{}H_2^{}:y\,mol\)

\(CH_4^{} + 2O_2^{}\xrightarrow{{{t^o}}}CO_2^{} + 2H_2^{}O\)

\(\,x\,\, \to \,\,\,x\)

\(C_2^{}H_2^{} + \frac{5}{2}O_2^{}\xrightarrow{{{t^o}}}2CO_2^{} + H_2^{}O\)

\(\,y\,\,\, \to \frac{5}{2}y\)

\(n_{hh}^{} = x + y = 0,01\,mol\)

\(n_{O_2^{}}^{} = 2{\text{x}} + 2,5y = 0,0225\,mol\)

\( \to x = 0,005\,mol;\,y = 0,005\,mol\)

Đề đoạn sau không rõ tính gì, nhưng tới đây là ra hết rồi chỉ cần tính bước cuối thôi