Câu hỏi của Xuyen Phan

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Lê Ng Hải Anh · Accepted Answer

$C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O$

a, Ta có: $V_{C_2H_5OH}=\dfrac{8.10}{100}=0,8\left(l\right)=800\left(cm^3\right)$

$\Rightarrow m_{C_2H_5OH}=800.0,8=640\left(g\right)\Rightarrow n_{C_2H_5OH}=\dfrac{640}{46}=\dfrac{320}{23}\left(mol\right)$

Theo PT: $n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=\dfrac{320}{23}\left(mol\right)$

Mà: H = 92% $\Rightarrow n_{CH_3COOH\left(TT\right)}=\dfrac{320}{23}.92\%=12,8\left(mol\right)$

$\Rightarrow m_{CH_3COOH}=12,8.60=768\left(g\right)$

b, Ta có: $m_{ddgiam}=\dfrac{768}{4\%}=19200\left(g\right)$

Câu trả lời:

$C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O$

a, Ta có: $V_{C_2H_5OH}=\dfrac{8.10}{100}=0,8\left(l\right)=800\left(cm^3\right)$

$\Rightarrow m_{C_2H_5OH}=800.0,8=640\left(g\right)\Rightarrow n_{C_2H_5OH}=\dfrac{640}{46}=\dfrac{320}{23}\left(mol\right)$

Theo PT: $n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=\dfrac{320}{23}\left(mol\right)$

Mà: H = 92% $\Rightarrow n_{CH_3COOH\left(TT\right)}=\dfrac{320}{23}.92\%=12,8\left(mol\right)$

$\Rightarrow m_{CH_3COOH}=12,8.60=768\left(g\right)$

b, Ta có: $m_{ddgiam}=\dfrac{768}{4\%}=19200\left(g\right)$