Cái này sai đề rồi em, anh lập hệ pt mà bấm ra số mol âm

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

\(\Rightarrow\left\{{}\begin{matrix}24x+40y=17,6\\x=2y\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

a)\(m_{Mg}=0,4\cdot24=9,6g\)

   \(m_{Ca}=0,2\cdot40=8g\)

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

   \(2Ca+O_2\underrightarrow{t^o}2CaO\)

Từ hai pt: \(\Rightarrow\Sigma n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{1}{2}n_{Ca}=\dfrac{1}{2}\cdot0,4+\dfrac{1}{2}\cdot0,2=0,3mol\)

\(\Rightarrow m_{O_2}=0,3\cdot32=9,6g\)

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)

a) 

Có \(\left\{{}\begin{matrix}24.n_{Mg}+40.n_{Ca}=17,6\\n_{Mg}=2.n_{Ca}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{Ca}=0,2\left(mol\right)\\n_{Mg}=0,4\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Ca}=0,2.40=8\left(g\right)\\m_{Mg}=0,4.24=9,6\left(g\right)\end{matrix}\right.\)

b)

PTHH: 2Ca + O₂ --t^o--> 2CaO 

            0,2-->0,1

             2Mg + O₂ --t^o--> 2MgO

            0,4--->0,2

=> \(V_{O_2}=\left(0,1+0,2\right).22,4=6,72\left(l\right)\)

\(V_{kk}=6,72.5=33,6\left(l\right)\)

\(n_{O_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(a.......\dfrac{2a}{3}\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

\(b.......\dfrac{3b}{4}\)

\(n_{O_2}=\dfrac{2a}{3}+\dfrac{3b}{4}=0.25\left(mol\right)\left(1\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{H_2}=a+1.5b=0.45\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.15,b=0.2\)

\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)

\(\%m_{Fe}=\dfrac{8.4}{8.4+5.4}\cdot100\%=60.8\%\)

\(\%m_{Al}=100-60.8=39.2\%\)

\(m_{tăng}=m_{O_2}=7.2\left(g\right)\)

\(n_{O_2}=\dfrac{7.2}{32}=0.225\left(mol\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.225\cdot22.4=25.2\left(l\right)\)

\(Đặt:n_{Mg}a\left(mol\right),n_{Cu}=b\left(mol\right),n_{Al}=c\left(mol\right)\)

\(Mg+\dfrac{1}{2}O_2\underrightarrow{t^0}MgO\)

\(Cu+\dfrac{1}{2}O_2\underrightarrow{t^0}CuO\)

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(TC:n_{O_2}=0.5a=0.5b=0.75c=\dfrac{0.225}{3}=0.075\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=0.15\\b=0.15\\c=0.1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0.15\cdot24=3.6\left(g\right)\\m_{Cu}=0.15\cdot64=9.6\left(g\right)\\m_{Al}=0.1\cdot27=2.7\left(g\right)\end{matrix}\right.\)

a+b) 2Zn+O2--->2ZnO

x------1/2x--------x

4Al+3O2---->2Al2O3

y--------0,75y----0,5y

Ta có

n O2=8,96/22,4=0,4(mol)

Ta có hệ pt

\(\left\{{}\begin{matrix}65x+27y=23,8\\0,5x+0,75y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)

%m Zn=0,2.65/23,8.100%=55%

%m Al=100-55=45%

c)m ZnO=0,2.81=16,2(g)

m Al2O3=102.0,2=20,4(g)

%m ZnO=\(\frac{16,2}{16,2+10,4}.100\%=44\%\)

%m Al2O3=100-44=64%

`4Al + 3O_2` $\xrightarrow{t^o}$ `2Al_2 O_3`

`0,2`    `0,15`                                `(mol)`

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

`1,5`     `0,75`                               `(mol)`

`n_[Al]=[5,4]/27=0,2(mol)`

`n_[O_2]=[16,8]/[22,4]=0,75(mol)`

 `=>m_[hh]=0,2.27+1,5.24=41,4(g)`

`=>%m_[Al]=[5,4]/[41,4].100~~13,04%`

`=>%m_[Mg]~~100-13,04~~86,96%`

no2 = 3,36 /22,4 = 0,15mol 

gọi x,y lần lượt là mol của Mg và Cu 

lập pthh của pu 

2Mg + O2 ----------->2MgO  (có nhiệt độ ) (1)

2mol    1mol              2mol 

x mol            1/2x mol

2Cu + O2 ------------>2CuO (có nhiệt độ ) (2) 

2mol    1mol               2mol 

y mol   1/2y mol    

m hỗn hợp =  mMg  + mCu 

11,2              24x       64y  (3)

tổng số mol O2 = nO2 (1) + nO2 (2)   

                 0,15  =    1/2x   + 1/2 y   (4)

từ 3,4 ta có hệ pt  11,2 = 24x  +     64y

                               0.15  = 1/2 x  + 1/2 y

   giải hệ pt ta được : x =0,2  y= 0,1

mMg = 0,2 . 24 =4,8 g

%mMg = (4,8 /11,2).100% =42,85% 

%mCu = 100% - 42,85% =57,15%