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

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

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

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

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

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

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

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

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

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

a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)

\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)

\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)

b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)

\(\rightarrow V_{kk}=4,48.5=11,2l\)

c. Có \(n_{Mg}=2n_{O_2}=0,2l\)

\(\rightarrow m_{Mg}=0,2.24=4,8g\)

\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)

\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)

a)

2Mg + O₂ --t^o--> 2MgO

2Zn + O₂ --t^o--> 2ZnO

b)

Gọi số mol Mg, Zn là a, b (mol)

=> 24a + 65b = 23,3 (1)

PTHH: 2Mg + O₂ --t^o--> 2MgO

               a-->0,5a------>a

            2Zn + O₂ --t^o--> 2ZnO

               b-->0,5b------>b

=> 40a + 81b = 36,1 (2)

(1)(2) => a = 0,7 (mol); b = 0,1 (mol)

\(n_{O_2}=0,5a+0,5b=0,4\left(mol\right)\)

=> \(V_{O_2}=0,4.22,4=8,96\left(l\right)\)

c) 

m_Mg = 0,7.24 = 16,8 (g)

m_Zn = 0,1.65 = 6,5 (g)

THAM KHẢO:

Gọi số mol Mg và Zn lần lượt là x, y

Ta có 24x + 65y=23.3

     40x + 81y=36.1

=) x=0.7

    y= 0.1

b)

c)

\(a,2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ 2Zn+O_2\rightarrow\left(t^o\right)2ZnO\\ b,n_{O_2}=\dfrac{36,1-23,3}{32}=0,4\left(mol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,4.22,4=8,96\left(l\right)\\ \text{Đ}\text{ặt}:a=n_{Mg};b=n_{Zn}\left(a,b>0\right)\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}24a+65b=23,3\\0,5a+0,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,7\\b=0,1\end{matrix}\right.\\ c,m_{Mg}=24a=24.0,7=16,8\left(g\right)\\ m_{Zn}=65b=65.0,1=6,5\left(g\right)\)

4Al + 3O2 => 2Al2O3

2Mg + O2 => 2MgO

giải hệ 27x+24y= 15,6

0,75x+0,5y= 0,4

=> x= 0,4 ;y= 0,2

=> mAl = 0,4.27 = 10,8(g)

%Al = \(\frac{10,8}{15,6}.100\%=69,23\%\)

=> %Mg = 100%- 69,23% = 30,77%

TN1: Gọi (n_Cu, n_Al, n_Fe) = (a,b,c)

=> 64a + 27b + 56c = 14,3 (1)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             b----------------------->1,5b

            Fe + 2HCl --> FeCl₂ + H₂

              c----------------------->c

=> 1,5b + c = 0,3 (2)

TN2: Gọi (n_Cu, n_Al, n_Fe) = (ak,bk,ck)

=> ak + bk + ck = 0,6 (3)

\(n_{O_2}=\dfrac{44,8}{22,4}.20\%=0,4\left(mol\right)\)

PTHH: 2Cu + O₂ --t^o--> 2CuO

            ak--->0,5ak

            4Al + 3O₂ --t^o--> 2Al₂O₃

           bk--->0,75bk

            3Fe + 2O₂ --t^o--> Fe₃O₄

           ck-->\(\dfrac{2}{3}ck\)

=> 0,5ak + 0,75bk + \(\dfrac{2}{3}ck\) = 0,4 (4)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,1\left(mol\right)\\c=0,15\left(mol\right)\\k=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,05.64}{14,3}.100\%=22,38\%\\\%m_{Al}=\dfrac{0,1.27}{14,3}.100\%=18,88\%\\\%m_{Fe}=\dfrac{0,15.56}{14,3}.100\%=58,74\%\end{matrix}\right.\)

Theo ĐLBT KL, có: m_KL + m_O2 = m _oxit

⇒ m_O2 = 28,4 - 15,6 = 12,8 (g)

\(\Rightarrow n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)

Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

Theo đề: \(m_{hh}=36\left(g\right)\)

\(\Rightarrow m_{Mg}+m_{Fe}=36\\ \Rightarrow24x+56y=36\left(1\right)\)

\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)....x\rightarrow...0,5x.....x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow...\dfrac{2}{3}y....\dfrac{1}{3}y\)

Theo đề: \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(\Rightarrow0,5x+\dfrac{2}{3}y=0,6\left(2\right)\)

\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}24x+56y=36\\0,5x+\dfrac{2}{3}y=0,6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,8.24=19,2\left(g\right)\\m_{Fe}=0,3.56=16,8\left(g\right)\end{matrix}\right.\\ m_r=m_{MgO}+m_{Fe_3O_4}=0,8.40+\dfrac{1}{3}.0,3.232=55,2\left(g\right)\)

PTHH: 

mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)

VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)

mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)

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

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

 x                              1/2 x         ( mol )

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

 y                               y     ( mol )

Ta có:

\(\left\{{}\begin{matrix}27x+64y=18,2\\51x+80y=26,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(\Rightarrow m_{Al}=0,2.27=5,4g\)

\(\Rightarrow m_{Cu}=0,2.64=12,8\)

\(\%m_{Al}=\dfrac{5,4}{18,2}.100=29,67\%\)

\(\%m_{Cu}=100\%-29,67=70,33\%\)