Ta có: `m1 = m2 = 11,05 (g)`

Phần 1:

PT: \(2Zn+O_2\underrightarrow{t^o}2ZnO\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

Theo ĐLBT KL, có: `mKL + mO2 = m` oxit

`⇒ mO2 = 18,25 - 11,05 = 7,2 (g)`

\(\Rightarrow n_{O_2}=\dfrac{7,2}{32}=0,225\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}+\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Mg}=0,225\left(mol\right)\)

\(\Rightarrow n_{Zn}+\dfrac{3}{2}n_{Al}+n_{Mg}=0,45\left(1\right)\)

Phần 2:

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}+n_{Mg}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow n_{H_2}=0,45\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)

Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,45\left(mol\right)\)

Theo ĐLBT KL, có: `mKL + mH2SO4 = m muối + mH2`

`⇒ m` chất rắn khan = `m` muối `= 11,05 + 0,45.98 - 0,45.2 = 54,25 (g)`

Giải thích các bước giải:

Khối lượng X mỗi phần là $\dfrac{22,1}2=11,05\ (gam)$

Phần 1:

$2Zn+O_2\xrightarrow{t^\circ} 2ZnO$

$4Al + 3O_2 \xrightarrow{t^\circ}  2Al_2O_3$

$2Mg+ O_2 \xrightarrow{t^\circ}  2MgO$

BTKL: $m_{KL}+m_O=m_{oxit}$

$⇒m_O=18,25-11,05=7,2\ (gam)$

$⇒n_O=\dfrac{7,2}{16}=0,45\ (mol)$

Phần 2:

$Zn+H_2SO_4\to ZnSO_4+H_2$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$Mg+H_2SO_4\to MgSO_4+H_2$

Ta xem xảy ra phản ứng:

$2H+O→H_2O$

$⇒n_H=0,45.2=0,9\ (mol)$

$⇒n_{H_2SO_4}=0,45\ (mol)$
BTNT H: $n_{H_2}=n_{H_2SO_4}=0,45\ (mol)$

$⇒V=0,45.22,4=10,08\ (l)$

$m=m_{KL}+m_{SO_4}=11,05+0,45.96=54,25\ (gam)$