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a) \(n_X=\dfrac{33,6}{22,4}=1,5\left(mol\right);n_{C_2Ag_2}=\dfrac{144}{240}=0,4\left(mol\right);n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH:
\(C_2H_2+2AgNO_3+2NH_3\rightarrow C_2Ag_2\downarrow+2NH_4NO_3\)
0,4<---------------------------------0,4
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b) \(n_{C_2H_4}=1,5-0,4-0,5=0,6\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CH_4}=0,5.16=8\left(g\right)\\m_{C_2H_2}=0,4.26=10,4\left(g\right)\\m_{C_2H_4}=0,6.28=16,8\left(g\right)\end{matrix}\right.\)
c) \(2CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\)
0,4------------------>0,2
\(\Rightarrow m_{C_4H_4}=0,2.80\%.52=8,32\left(g\right)\)
\(\text{Đ}\text{ặt}:n_{Al}=a\left(mol\right);n_{Cu}=b\left(mol\right)\left(a,b>0\right)\\ Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\\ Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}27a+64b=7,75\\3.22,4a+2.22,4b=7,84\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\\ a,\Rightarrow\%m_{Al}=\dfrac{0,05.27}{7,75}.100\approx17,419\%\\ \Rightarrow\%m_{Cu}\approx82,581\%\\ b,n_{HNO_3}=6a+4b=0,7\left(mol\right)\\ C_{M\text{dd}HNO_3}=\dfrac{0,7}{0,14}=5\left(M\right)\)
a) PTHH: Al + 6 HNO3 -> Al(NO3)3 + 3 NO2 + 3 H2O
x___________6x________x________3x(mol)
Fe + 6 HNO3 -> Fe(NO3)3 + 3 NO2 + 3 H2O
y___6y______y____________3y(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}27x+56y=1,95\\22,4.3x+22,4.3y=2,688\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,01\\y=0,03\end{matrix}\right.\)
b) Khối lượng mỗi kim loại trong hỗn hợp ban đầu:
mAl=27x=27 . 0,01=0,27(g)
mFe=56y= 56 . 0,03= 1,68(g)
c) m=m(muối)=mAl(NO3)3 + mFe(NO3)3= 213x+242y=213.0,01+ 242.0,03=9,39(g)
Gọi $n_{Fe} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 56a + 65b = 1,77(1)$
$n_{NO_2} = \dfrac{1,792}{22,4} = 0,08(mol)$
Bảo toàn electron :
$3n_{Fe} + 2n_{Zn} = n_{NO_2} \Rightarrow 3a + 2b = 0,08(2)$
Từ (1)(2) suy ra: a = 0,02 ; b = 0,01
$\%m_{Fe} = \dfrac{0,02.56}{1,77}.100\% = 63,3\%$
$\%m_{Zn} =100\% - 63,3\% = 36,7\%$
2.
Ta có Fe và Al thụ động với HNO3đặc nguội nên chỉ có Cu Phản ứng
Phần 1 : BTe\(\rightarrow\) nCu = 0,1 mol
Phần 2:
Đặt nAl = a ; nFe = b
BTKL\(\rightarrow\)27a + 56b = 11
BTe \(\rightarrow\) 3a + 2b = 0,8
\(\rightarrow\)a= 0,2 ; b = 0,1
\(\rightarrow\)mAl = 10,8 ; mFe = 11,2 gam
a, PT: \(C_2H_2+AgNO_3+NH_3\rightarrow Ag_2C_{2\downarrow}+NH_4NO_3\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, Ta có: \(V_{CH_4}=V_Y=0,56\left(mol\right)\Rightarrow\%V_{CH_4}=\dfrac{0,56}{6,72}.100\%\approx8,33\%\)
\(V_{C_2H_4}=V_X-V_{CH_4}=3,92\left(l\right)\Rightarrow\%V_{C_2H_4}=\dfrac{3,92}{6,72}.100\%\approx58,34\left(\%\right)\)
\(\Rightarrow V_{C_2H_2}=6,72-4,48=2,24\left(l\right)\Rightarrow\%V_{C_2H_2}=\dfrac{2,24}{6,72}.100\%\approx33,33\%\)
c, \(n_{Ag_2C_2}=n_{C_2H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Ag_2C_2}=0,1.240=24\left(g\right)\)
1)
Zn + 2HCl --> ZnCl2 + H2
2Al + 6HCl --> 2AlCl3 + 3H2
Zn + 4HNO3 --> Zn(NO3)2 + 2NO2 + 2H2O
2)
TN2:
\(n_{NO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Zn + 4HNO3 --> Zn(NO3)2 + 2NO2 + 2H2O
_____0,05<--------------------------0,1
=> nZn = 0,05 (mol)
TN1:
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
____0,05--------------------->0,05
2Al + 6HCl --> 2AlCl3 + 3H2
0,1<-----------------------0,15
=> m = 0,05.65 + 0,1.27 = 5,95(g)