Giúp mình với, hứa vote, cảm ơn nhiều

Tham khảo :

$a,$

$\mathop{C_2}\limits^{-2}H_5OH + K_2\mathop{Cr_2}\limits^{+6}O_7 + H_2SO_4 \xrightarrow{Ag^+} \mathop{C}\limits^{-3}H_3\mathop{C}\limits^{+3}OOH + K_2SO_4 + \mathop{Cr_2}\limits^{+3}(SO_4)_3 + H_2O$

$\mathop{C}\limits^{-2} \to \mathop{C}\limits^{+3}+5e^{-}| $$\rm x3$

$\mathop{C}\limits^{-2}+e^{-} \to \mathop{C}\limits^{-3}+$$\rm x3$

$\mathop{2Cr}\limits^{+6}+6e^{-} \to \mathop{2Cr}\limits^{+3}|$$\rm x 2$

`->`$3C_2H_5OH+2K_2Cr_2O_7+8H_2SO_4 \xrightarrow{Ag^{+}}3CH_3COOH+2 K_2SO_4 + 2Cr_2(SO_4)_3 + 11H_2O$

$b,$

$n_{K_2Cr_2O_7}=\dfrac{0,056.2}{294}.\dfrac{mg.ml.mmol}{ml.mg}=\dfrac{1}{2625}(mmol)$

$n_{C_2H_5OH}=\dfrac{3}{2}n_{K_2Cr_2O_7}=\dfrac{3}{2}.\dfrac{1}{2625}=\dfrac{1}{1750}(mmol)$

$m_{C_2H_5OH}=\dfrac{1}{1750}.46.\dfrac{mmol.mg}{mmol}=\dfrac{23}{875}(mg)$

$\dfrac{m_{C_2H_5OH}}{V_{hơi \,\, thở}}=\dfrac{\dfrac{23}{875}}{52,5}.\dfrac{1000}{1}.\dfrac{mg.ml}{ml.l}=\dfrac{368}{735}(mg/l)≈0,501(mg/l)$

Người này có vi phạm pháp luật và mức đóng phạt là $6000000-8000000$ đồng.