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

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

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

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

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

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

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

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

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

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

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

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

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

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% 

giúp tui với tui cần gấp

\(n_{CO_2}=\dfrac{8.8}{44}=0.2\left(mol\right)\)

\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)

\(2CO+O_2\underrightarrow{t^0}2CO_2\)

\(0.2.......0.1.......0.2\)

\(2H_2+O_2\underrightarrow{t^0}2H_2O\)

\(0.4......0.3-0.1\)

\(\%m_{CO}=\dfrac{0.2\cdot28}{0.2\cdot28+0.4\cdot2}\cdot100\%=87.5\%\)

\(\%m_{H_2}=100-87.5=12.5\%\)

a)

\(2CO + O_2 \xrightarrow{t^o} 2CO_2(1)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O(2) \)

b)

\(n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)\)

Theo PTHH :

\(n_{CO} = n_{CO_2} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{1}{2}n_{CO_2} = 0,1(mol)\\ n_{H_2} = 2n_{O_2(2)} = 2(0,3-0,1) = 0,4(mol)\)

Vậy :

\(\%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{13,44}{22,4}=0,6mol\)

Gọi \(\left\{{}\begin{matrix}n_P=x\\n_S=y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m_P=31x\\m_S=32y\end{matrix}\right.\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

 x      1,25x                      ( mol )

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

 y       y              ( mol )

Ta có:

\(\left\{{}\begin{matrix}31x+32y=15,6\\1,25x+y=0,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,4\\y=0,1\end{matrix}\right.\)

\(\Rightarrow m_P=31.0,4=12,4g\)

\(\Rightarrow m_S=32.0,1=3,2g\)

D

D