a) \(\text{Đặt }n_{Cl_2}=x\left(mol\right);n_{O_2}=y\left(mol\right)\)

\(\Rightarrow n_{kh\text{í}}=x+y=0,5\left(1\right)\)

\(BTKL\Rightarrow m_{kh\text{í}}=m_Z-m_{KL}=25,36\left(g\right)\\ \Rightarrow71x+32y=25,36\left(2\right)\)

Từ (1) và \(\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,24\\y=0,26\end{matrix}\right.\)

\(\Rightarrow\%V_{O_2}=52\%\)

b) \(\text{Đặt }n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\)

\(\Rightarrow m_{KL}=24x+27y=16,98\left(3\right)\)

\(BT\text{ }e\Rightarrow2n_{Mg}+3n_{Al}=2n_{Cl_2}+4n_{O_2}\\ \Rightarrow2a+3b=1,52\left(4\right)\)

Từ (3) và \(\left(4\right)\Rightarrow\left\{{}\begin{matrix}a=0,55\\b=0,14\end{matrix}\right.\)

\(\Rightarrow\%Mg=77,74\%\)

a)

\(\text{2Fe + O2}\)\(\underrightarrow{^{to}}2FeO\)

\(\text{3Fe + 2O2}\)\(\underrightarrow{^{to}}Fe3O4\)

\(\text{2Cu + O2}\underrightarrow{^{to}}2CuO\)

\(\text{4Al + 3O2}\underrightarrow{^{to}}2Al2O3\)

b)

mO2 p.ứ = \(\frac{8.20}{100}\) = 1,6 (g)
Bảo toàn khối lượng: mA + mO2 p.ứ = mX

→ mA = 24,05 - 1,6 = 22,45 (g)

1

a)1:8:2:3:4

b)

c)1:(3n-1):n:(n-1)

d)1:8:1:2:4

e)2:2n:2:n

mik viết theo tỉ lệ hệ số nha!!!

good luck!!!

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

\(n_{H2}=0,12\left(mol\right)\Rightarrow n_{Al}=0,08\left(mol\right)\)

\(\Rightarrow m_{Al}=0,08.27=2,16\left(g\right)\)

PTPU

CuO+ H₂\(\xrightarrow[]{to}\) Cu+ H₂O

CuO+ CO\(\xrightarrow[]{to}\) Cu+ CO₂

Fe₃O₄+ 4H₂\(\xrightarrow[]{to}\) 3Fe+ 4H₂O

Fe₃O₄+ 4CO\(\xrightarrow[]{to}\) 3Fe+ 4CO₂

hỗn hợp khí sau phản ứng là CO₂ và H₂O

ta có: m_{CO2+ H2O}= m_{CO+ H2}+ m_{O( CuO, Fe3O4)}

= m_{CO+ H2}+ 0,16

m_{hh oxit}= m_Cu+ m_Fe+ m_Al2O3+ m_{O(CuO, Fe3O4)}

a= m_Cu+ m_Fe+ m_Al2O3

\(\Rightarrow\) a= m_{hh oxit}- m_O(CuO,Fe3O4)

= 8,4- 0,16= 8,24(g)

n_O(CuO,Fe3O4)= \(\dfrac{0,16}{16}\)= 0,01( mol)

có: n_O(CuO,Fe3O4)= n_CO2+ n_H2

V_CO2+ V_H2= 22,4.( n_CO2+ n_H2)

= 22,4. n_O(CuO,Fe3O4)

= 22,4. 0,1= 2,24( lít)

PTPU

CuO+ H₂\(\xrightarrow[]{to}\) Cu+ H₂O

CuO+ CO\(\xrightarrow[]{to}\) Cu+ CO₂

Fe₃O₄+ 4H₂\(\xrightarrow[]{to}\) 3Fe+ 4H₂O

Fe₃O₄+ 4CO\(\xrightarrow[]{to}\) 3Fe+ 4CO₂

hồn hợp khí sau phản ứng là CO₂ và H₂O

ta có: m_{hh khí sau}= m_{hh khí ban đầu}+ m_{O( CuO, Fe3O4)}

= m_{hh khí ban đầu}+ 0,16

m_{hh oxit ban đầu}= m_CuO+ m_Fe3O4+ m_Al2O3

= m_Cu+ m_Fe+ m_Al2O3+ m_{O( CuO, Fe3O4)}

a= m_Cu+ m_Fe+ m_Al2O3

\(\Rightarrow\) a= m_{hh oxit ban đầu}- m_{O( CuO, Fe3O4)}

= 8,4- 0,16=8,24( g)

có: n_{O( CuO, Fe3O4)}= \(\dfrac{0,16}{16}\)= 0,01( mol)

n_{O( CuO, Fe3O4)}= n_CO2+ n_H2

\(\Rightarrow\) V_{hh khí ban đầu}= 22,4.( n_CO2+ n_H2)

= 22,4. n_{O( CuO, Fe3O4)}

= 22,4. 0,01= 0,224( lít)