Các CTCT phù hợp của (X) là:

\(CH_3C\text{OO}CH_3\\ HCOOC_2H_5\\ C_2H_5C\text{OO}H\)

=>C

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2 +H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\ a,V_{H_2\left(\text{đ}ktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,m_{FeCl_2}=127.0,15=19,05\left(mol\right)\\ m_{\text{dd}FeCl_2}=m_{Fe}+m_{\text{dd}HCl}-m_{H_2}=8,4+200-0,15.2=208,1\left(g\right)\\ C\%_{\text{dd}FeCl_2}=\dfrac{19,05}{208,1}.100\approx9,154\%\)

Đổi: 12,6dm²=1260cm²

Cạnh đáy dài:

(1260 x 2): 45 =56(cm)

Em xem lại đề nha em!

\(n_{O_2}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)

=> C9: C

\(n_{Fe_3O_4}=\dfrac{0,15}{3}=0,05\left(mol\right)\\ m_{Fe_3O_4}=232.0,05=11,6\left(g\right)\)

=> C10: B

2 : dị dưỡng, tự dưỡng