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Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Na}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
a--------------------------->a
\(2Na+2HCl\rightarrow2NaCl+H_2\)
b---------------------------->0,5b
Ta có: \(m_M=\dfrac{1}{2}.\left(m_{Fe}+m_{Na}\right)=\dfrac{1}{2}.\left(56a+23b\right)=28a+11,5b\left(g\right)\)
PTHH: \(M+2HCl\rightarrow MCl_2+H_2\)
(a+0,5b)<----------------(a+0,5b)
\(\Rightarrow M_M=\dfrac{28a+11,5b}{a+0,5b}\\ \Rightarrow\dfrac{28a}{a}>M_M>\dfrac{11,5a}{0,5a}\\ \Leftrightarrow28>M_M>23\)
Vậy M là Magie (Mg)
Bài 5:
mCu= 43,24% . 14,8\(\approx\) 6,4(g)
=>mFe\(\approx\) 14,8 - 6,4= 8,4(g)
=> nFe\(\approx\) 8,4/56\(\approx\) 0,15(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
nH2=nFe \(\approx\) 0,15 (mol)
=> V(H2,đktc) \(\approx\) 0,15 . 22,4\(\approx\) 3,36(l)
Bài 6:
nH2= 4,368/22,4=0,195(mol)
Đặt: nMg=a(mol); nAl=b(mol) (a,b>0)
PTHH: Mg + 2 HCl -> MgCl2 + H2
a________2a_____a_____a(mol)
2 Al + 6 HCl -> 2 AlCl3 +3 H2
b____3b____b______1,5b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+27b=3,87\\a+1,5b=0,195\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,06\\b=0,09\end{matrix}\right.\)
a) nH2SO4= 2a+3b=0,39(mol)
=> mH2SO4= 0,39.98=38,22(g)
b) m(muối)= mMgSO4 + mAl2(SO4)3= 120a+ 133,5b= 120.0,06+133,5.0,09= 19,215(g)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,2------------>0,2----->0,2
=> \(\left\{{}\begin{matrix}m_{FeCl_2}=0,2.127=25,4\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)
a)
Gọi số mol Fe, Mg, Al là a, b,c (mol)
=> 56a + 24b + 27c = 6,4 (1)
nHCl = 0,5.1,6 = 0,8 (mol)
PTHH: Fe + 2HCl --> FeCl2 + H2
a--->2a-------------->a
Mg + 2HCl --> MgCl2 + H2
b----->2b------------->b
2Al + 6HCl --> 2AlCl3 + 3H2
c-->3c---------------->1,5c
=>nHCl(pư)=2a+2b+3c= \(\dfrac{56a}{28}+\dfrac{24b}{12}+\dfrac{27c}{9}< \dfrac{56a+24b+27c}{9}=\dfrac{6,4}{9}< 0,8\)
=> A tan hết
b)
\(n_{CuO}=\dfrac{18,4}{80}=0,23\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
0,23->0,23
=> a + b + 1,5c = 0,23 (2)
\(m_{Al}=\dfrac{6,4.33,75}{100}=2,16\left(g\right)\)
=> \(c=\dfrac{2,16}{27}=0,08\left(mol\right)\) (3)
(1)(2)(3) => a = 0,05 (mol); b = 0,06 (mol)
=> \(\left\{{}\begin{matrix}m_{Al}=2,16\left(g\right)\\m_{Fe}=0,05.56=2,8\left(g\right)\\m_{Mg}=0,06.24=1,44\left(g\right)\end{matrix}\right.\)
$a\big)$
$M_A=9,4.2=18,8(g/mol)$
$\to \dfrac{n_{CO_2}}{n_{H_2}}=\dfrac{18,8-2}{44-18,8}=\dfrac{2}{3}$
Mà $n_{CO_2}+n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)$
\(\begin{array} {l} \to n_{CO_2}=0,2(mol);n_{H_2}=0,3(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ FeCO_3+2HCl\to FeCl_2+CO_2+H_2O\\ \text{Theo PT: }n_{Fe}=n_{H_2}=0,3(mol);n_{FeCO_3}=n_{CO_2}=0,2(mol)\\ \to m=0,3.56+0,2.116=40(g) \end{array}\)
$b\big)$
Đổi $400ml=0,4l$
\(\begin{array} {l} \text{Theo PT: }n_{FeCl_2}=n_{H_2}+n_{CO_2}=0,5(mol)\\ \to C_{M\,FeCl_2}=\dfrac{0,5}{0,4}=1,25M \end{array}\)
$c\big)$
\(\begin{array}{l} m_{dd\,FeCl_2}=\dfrac{400}{1,2}\approx 333,33(g)\\ \to C\%_{FeCl_2}=\dfrac{0,5.127}{333,33}.100\%=19,05\%\end{array}\)
Giả sử 21,4 gam X chứa \(\left\{{}\begin{matrix}Cu:a\left(mol\right)\\Fe:2a\left(mol\right)\\R:b\left(mol\right)\end{matrix}\right.\)
=> 176a + b.MR = 21,4 (1)
\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
2a--------------------->2a
2R + 2nHCl --> 2RCln + nH2
b---------------------->0,5bn
=> 2a + 0,5bn = 0,7 (2)
- 10,7g X chứa \(\left\{{}\begin{matrix}Cu:0,5a\left(mol\right)\\Fe:a\left(mol\right)\\R:0,5b\left(mol\right)\end{matrix}\right.\)
\(n_{Cl_2}=\dfrac{39,1-10,7}{71}=0,4\left(mol\right)\)
PTHH: Cu + Cl2 --to--> CuCl2
0,5a->0,5a
2Fe + 3Cl2 --to--> 2FeCl3
a-->1,5a
2R + nCl2 --to--> 2RCln
0,5b->0,25bn
=> 2a + 0,25bn = 0,4 (3)
(2)(3) => a = 0,05 (mol); bn = 1,2
=> \(\left\{{}\begin{matrix}n_{Cu}=0,05\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,05.64}{21,4}.100\%=14,95\%\\\%m_{Fe}=\dfrac{0,1.56}{21,4}.100\%=26,17\%\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=0,1.1,2=0,12\left(mol\right)\\ n_{H_2}=0,05\left(mol\right)\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2
a 3a a 1,5a
Fe + 2HCl ---> FeCl2 + H2
b 2b b b
Hệ pt \(\left\{{}\begin{matrix}27a+56b=1,66\\1,5a+b=0,05\end{matrix}\right.\Leftrightarrow a=b=0,02\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,02.27=0,54\left(g\right)\\m_{Fe}=0,02.56=1,12\left(g\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,02}{0,1}=0,2M\\C_{M\left(FeCl_2\right)}=\dfrac{0,02}{0,1}=0,2M\\C_{M\left(HCl.dư\right)}=\dfrac{0,12-0,02.3-0,02.2}{0,1}=0,2M\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,47}{22,4}\approx0,21\left(mol\right)\\ PTHH:\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Theo các pthh:
\(n_{H_2SO_4}=n_{H_2}=0,21\left(mol\right)\\ \rightarrow m_{H_2SO_4}=0,21.98=20,58\left(g\right)\\ \rightarrow m_{ddH_2SO_4}=\dfrac{20,58}{9,8\%}=210\left(g\right)\\ m_{H_2}=0,21.2=0,42\left(g\right)\\ \rightarrow m_{dd\left(sau\right)}=210+4,46-0,42=214,04\left(g\right)\)
\(\left\{{}\begin{matrix}m_{Mg}=\dfrac{40.9}{100}=3,6\left(g\right)\\m_{Al}=9-3,6=5,4\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\end{matrix}\right.\)
PTHH:
Mg + 2HCl ---> MgCl2 + H2
0,15 ------------------------> 0,15
2Al + 6HCl ---> 2AlCl3 + 3H2
0,2 ---------------------------> 0,3
\(\rightarrow V_{H_2}=\left(0,15+0,3\right).22,4=10,08\left(l\right)\)
\(n_{H_2O}=\dfrac{6,48}{18}=0,36\left(mol\right)\)
PTHH: FexOy + yH2 --to--> xFe + yH2O
Theo pthh: \(n_{O\left(oxit\right)}=n_{H_2\left(pư\right)}=n_{H_2O}=0,36\left(mol\right)\)
\(\rightarrow m_{oxit}=15,12+16.0,36=20,88\left(g\right)\)
\(n_{Fe}=\dfrac{15,12}{56}=0,27\left(mol\right)\)
CTHH: FexOy
=> x : y = 0,27 : 0,36 = 3 : 4
=> CTHH: Fe3O4 (oxit sắt từ)
mMg = 40%x9 = 3,6(g) =>nMg=3,6:24 = 0,15 (mol)
=> mAl = 9-3,6 = 5,4(g) => nAl = 5,4:27 = 0,2 (mol)
pthh : 2Al+6HCl -> 2AlCl3+3H2
0,2 0,3
Mg+2HCl -> MgCl2 +H2
0,15 0,15
=> nH2 = 0,15 + 0,3 = 0,45 (mol)
=> VH2 = 0,45.22,4 = 10,08 (L)
mH2 = 0,45 . 2 = 0,9 (mol)
áp dụng BLBTKL ta có :
mH2 + moxit sắt = mFe + mH2O
=> moxit sắt = 20,7 (g)
\(m_{Fe}=16.70\%=11,2\left(g\right)\\ m_{Mg}=16-11,2=4,8\left(g\right)\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH:
`Fe + 2HCl -> FeCl_2 + H_2`
0,2------------------------>0,2
`Mg + 2HCl -> MgCl_2 + H_2`
0,2------------------------->0,2
`V = 22,4.(0,2 + 0,2) = 8,96 (l)`
\(m_{Fe}=16.70\%=11,2\left(g\right)\)
\(\rightarrow m_{Mg}=16-11,2=4,8\left(g\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,2 0,2 ( mol )
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,2 0,2 ( mol )
\(V_{H_2}=\left(0,2+0,2\right).22,4=8,96\left(l\right)\)