\(n_{hhkhí}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{ZnS}=b\left(mol\right)\end{matrix}\right.\)

Bảo toàn khối lượng: \(m_A=m_Y=35,6\left(g\right)\)

PTHH:

Zn + 2HCl ---> ZnCl₂ + H₂

a -------------------------> a

ZnS + 2HCl ---> ZnCl₂ + H₂S

b ------------------------------> b

Hệ pt \(\left\{{}\begin{matrix}65a+97b=35,6\\a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\)

PTHH: Zn + S --t^o--> ZnS

           0,3 <- 0,3 <----- 0,3

\(\rightarrow\left\{{}\begin{matrix}m_{Zn}=\left(0,1+0,3\right).65=26\left(g\right)\\m_S=0,3.32=9,6\left(g\right)\end{matrix}\right.\)

n SO2=\(\dfrac{38,528}{22,4}\) =1,72 mol

m ct NaOH=\(\dfrac{2832.10}{100}\) =283,2 g

n NaOH=\(\dfrac{283,2}{40}\) =7,08 mol

T=\(\dfrac{7,08}{1,72}\)≈4,1 >2

⇒NaOH dư, tính theo SO2

⇒tạo muối trung hòa

SO2+2NaOH→Na2SO3+H2O

1,72→                 1,72                       mol

m Na2SO3=1,72.126=216,72 g

ta có :

2Mg+O2-to>2MgO

x--------0,5x

2Zn+O2-to>2ZnO

y-------0,5y

=>\(\left\{{}\begin{matrix}24x+65y=14,58\\0,5x+0,5y=0,15\end{matrix}\right.\)

=>x=0,12 mol ,y=0,18 mol

=>%mMg=\(\dfrac{0,12.24}{14,58}100\)=19,753%

=>%mZn=80,247%

\(n_{O_2}=\dfrac{4,8}{32}=0,15mol\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Zn}=y\end{matrix}\right.\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

  x        1/2 x                       ( mol )

\(2Zn+O_2\rightarrow\left(t^o\right)2ZnO\)

  y       1/2 y                         ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+65y=14,58\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,12\\y=0,18\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,12.24=2,88g\\m_{Zn}=0,18.65=11,7g\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,88}{14,58}.100=19,75\%\\\%m_{Zn}=100\%-19,75\%=80,25\%\end{matrix}\right.\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ pthh:2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
         0,2                           0,1                 0,3 
\(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\V_{H_2}=0,3.22,4=6,72\left(L\right)\end{matrix}\right.\)

Giải thích các bước giải:

2Al+6H2SO4→Al2(SO4)3+3SO2+6H2O

nAl=5,4/27=0,2mol

nSO2=0,2×32=0,3mol

⇒VSO2=0,3×22,4=6,72l

m chx bt tính kiểu j

Đặt m_T = a (g)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2}=50\%.a=0,5a\left(g\right)\\m_{O_3}=a-0,5a=0,5a\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{0,5a}{32}=0,015625a\left(mol\right)\\n_{O_3}=\dfrac{0,5a}{48}=\dfrac{a}{96}\left(mol\right)\end{matrix}\right.\\ \rightarrow M_T=\dfrac{a}{0,015625a+\dfrac{a}{96}}=38,4\left(\dfrac{g}{mol}\right)\)

=> d_T/He = \(\dfrac{38,4}{4}=9,6\)

\(n_{FeS_2}=\dfrac{240}{120}.80\%=1,6\left(mol\right)\)

PTHH: 4FeS₂ + 11O₂ --t^o--> 2Fe₂O₃ + 8SO₂

               1,6 -------------------------------> 3,2

2SO₂ + O₂ --t^o--> 2SO₃

3,2 --------------------> 3,2

SO₃ + H₂O ---> H₂SO₄

3,2 ----------------> 3,2

\(m_{ddH_2SO_4}=\dfrac{3,2.98}{49\%}=640\left(g\right)\)

4FeS2+11O2-to>2Fe2O3+8SO2

1,6----------------------------------3,2

2SO2+O2-to,V2O5->2SO3

3,2----------------------------3,2

SO3+H2o->H2SO4

3,2-----------------3,2

n FeS2=2 mol

H=80%

=>n FeS2=2.80%=1,6 mol

=>m H2SO4=3,2.98=313,6g

=>mdd=640g

=>B