$23)$

$n_{Cl_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$\Rightarrow m_{Cl_2}=0,2.71=14,2(g)$

$\to B$

$24)$

$Cu+2H_2SO_{4(đ)}\to CuSO_4+SO_4\uparrow+2H_2O$

Tỉ lệ: $1:2:1:1:2$

$\to C$

$25)CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow$

$\Rightarrow m_{CaCO_3}=m_{CO_2}+m_{CaO}$

$\Rightarrow m_{CaO}<m_{CaCO_3}\Rightarrow m_{rắn}$ giảm

$\to A$

$26)$ Bảo toàn KL:

$m_X=m_{oxit}+m_{CO_2}$

$\Rightarrow m_{oxit}=31,8-15,4=16,4(g)$

$\to B$

$27)$

$PTHH:4FeS_2+11O_2\xrightarrow{t^o}2Fe_2O_3+8SO_2\uparrow$

$\Rightarrow x:y=4:11$

$\to A$

$28)$

$n_{Fe}=\dfrac{140}{56}=2,5(mol)$

$Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\uparrow$

Theo PT: $n_{CO}=1,5.n_{Fe}=3,75(mol)$

$\Rightarrow V_{CO(đktc)}=3,75.22,4=84(lít)$

$\to $ Không đáp án nào đúng

hay quá top 1 bình luận pặn lun :')

:') đc đc iu iem

$15)$

$n_{MgO}=\dfrac{24}{40}=0,6(mol)$

$\Rightarrow A_{MgO}=0,6.6.10^{23}=3,6.10^{23}$

$\to B$

$16)\%m_{Cu}=\dfrac{64}{64+32+16.4}.100\%=40\%$

$\to C$

$17)PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\uparrow$

$n_{Zn}=\dfrac{3,25}{65}=0,05(mol)$

Theo PT: $n_{H_2}=n_{Zn}=0,05(mol)$

$\Rightarrow V_{H_2}=0,05.22,4=1,12(lít)$

$\to B$

$18)PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5$

Theo PT: $n_{O_2}=1,25.n_P=1,875(mol)$

$\to D$

$19)$ A

K tương ứng x sao anh lại suy ra y?