Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
1. Gọi mol của Mg và Al là x, y mol
=> 24x + 27y = 12,6 (1)
nH2 = 0,6 mol => x + 1,5y = 0,6 (2)
Từ (1) (2) => x = 0,3 ; y = 0,2
=> %Mg = 57,14%
=> %Al = 42,86%
nH2=13,44/22,4=0,6(mol)
Đặt: nMg=a(mol); nAl=b(mol) (a,b>0)
1) PTHH: Mg + H2SO4 -> MgSO4 + H2
a__________a________a_____a(mol)
2Al +3 H2SO4 -> Al2(SO4)3 + 3 H2
b___1,5b______0,5b____1,5b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+27b=12,6\\a+1,5b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)
=> mMg=0,3.24=7,2(g)
=>%mMg= (7,2/12,6).100=57,143%
=>%mAl=42,857%
2) mMgSO4=120.a=120.0,3=36(g)
mAl2(SO4)3=342.0,5b=342.0,5.0,2= 34,2(g)
mH2SO4= (0,3+0,2.1,5).98=58,8(g)
=>mddH2SO4=58,8: 14,7%=400(g)
=>mddsau= 12,6+400 - 2.0,6= 411,4(g)
=>C%ddAl2(SO4)3= (34,2/411,4).100=8,313%
C%ddMgSO4=(36/411,4).100=8,751%
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
\(n_{H_2}=\frac{10,08}{22,4}=0,45\left(mol\right)\)
a) Theo pt: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,45=0,3\left(mol\right)\)
\(\Rightarrow m_{Al}=0,3\times27=8,1\left(g\right)\)
\(\Rightarrow m_{Cu}=14,5-8,1=6,4\left(g\right)\)
\(\%m_{Al}=\frac{8,1}{14,5}\times100\%=55,86\%\)
\(\%m_{Cu}=\frac{6,4}{14,5}\times100\%=44,14\%\)
b) Theo pT: \(n_{H_2SO_4}=n_{H_2}=0,45\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,45\times98=44,1\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{44,1}{20\%}=220,5\left(g\right)\)
c) \(m_{H_2}=0,45\times2=0,9\left(g\right)\)
Ta có: \(m_{dd}saupứ=8,1+220,5-0,9=227,7\left(g\right)\)
Theo pT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2}=\frac{1}{3}\times0,45=0,15\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,15\times342=51,3\left(g\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{51,3}{227,7}\times100\%=22,53\%\)
Gọi số mol của Al và Cu lần lượt là x và y
Vì Cu k phản ứng với dd H2SO4 20% ( đã bị pha loãng) nên ta có
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
(mol) 2 3 1 3
(mol) x 3x/2 x/2 3x/2
Theo đề bài ta có:
\(hpt:\left\{{}\begin{matrix}22,4\times\frac{3x}{2}=10,08\\27x+64y=14,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Al}=0,3\left(mol\right)\rightarrow\%m_{Al}=\frac{0,3.27}{14,5}.100\%=55,9\left(\%\right)\\n_{Cu}=0,1\left(mol\right)\rightarrow\%m_{Cu}=100-55,9=44,1\left(\%\right)\end{matrix}\right.\)
\(m_{H_2SO_4}=n.M=\frac{98.3x}{2}=98.\frac{3.0,3}{2}=44,1\left(g\right)\)
\(m_{ddH_2SO_4}=\frac{44,1.100\%}{20\%}=220,5\left(g\right)\)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\frac{m_{ct}}{m_{dd}}.100\%=\frac{342.\frac{0,3}{2}}{220,5+14,5-2.\frac{3.0,3}{2}}.100\%=21,91\left(\%\right)\)
a) Do dd sau pư có 3 chát tan với nồng độ % bằng nhau
=> \(m_{Al_2\left(SO_4\right)_3}=m_{ZnSO_4}=m_{H_2SO_4\left(dư\right)}\)
Gọi số mol Al, Zn là a, b (mol)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
a----->1,5a------->0,5a----->1,5a
Zn + H2SO4 --> ZnSO4 + H2
b----->b--------->b----->b
=> \(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=342.0,5a=171a\left(g\right)\\m_{ZnSO_4}=161b\left(g\right)\end{matrix}\right.\)
=> 171a = 161b
=> \(\dfrac{a}{b}=\dfrac{161}{171}\) (1)
Có: \(\dfrac{m_{Al}}{m_{Zn}}=\dfrac{27.n_{Al}}{65.n_{Zn}}=\dfrac{27}{65}.\dfrac{161}{171}=\dfrac{483}{1235}\)
b) \(n_{H_2}=1,5a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) (2)
(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{161}{825}\left(mol\right)\\b=\dfrac{57}{275}\left(mol\right)\end{matrix}\right.\)
=> \(x=\dfrac{161}{825}.27+\dfrac{57}{275}.65=\dfrac{5154}{275}\left(g\right)\)
\(m_{H_2SO_4\left(dư\right)}=m_{Al_2\left(SO_4\right)_3}=342.0,5\dfrac{161}{825}=\dfrac{9177}{275}\left(g\right)\)
=> \(m_{H_2SO_4\left(bđ\right)}=98\left(1,5a+b\right)+\dfrac{9177}{275}=\dfrac{22652}{275}\left(g\right)\)
=> \(y=\dfrac{\dfrac{22652}{275}.100}{10}=\dfrac{45304}{55}\left(g\right)\)
a) Do dd sau pư có 3 chát tan với nồng độ % bằng nhau
=> \(m_{Al_2\left(SO_4\right)_3}=m_{ZnSO_4}=m_{H_2SO_4\left(dư\right)}\)
Gọi số mol Al, Zn là a, b (mol)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
a----->1,5a------->0,5a----->1,5a
Zn + H2SO4 --> ZnSO4 + H2
b----->b--------->b----->b
=> \(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=342.0,5a=171a\left(g\right)\\m_{ZnSO_4}=161b\left(g\right)\end{matrix}\right.\)
=> 171a = 161b
=> \(\dfrac{a}{b}=\dfrac{161}{171}\) (1)
Có: \(\dfrac{m_{Al}}{m_{Zn}}=\dfrac{27.n_{Al}}{65.n_{Zn}}=\dfrac{27}{65}.\dfrac{161}{171}=\dfrac{483}{1235}\)
b) \(n_{H_2}=1,5a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) (2)
(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{161}{825}\left(mol\right)\\b=\dfrac{57}{275}\left(mol\right)\end{matrix}\right.\)
=> \(x=\dfrac{161}{825}.27+\dfrac{57}{275}.65=\dfrac{5154}{275}\left(g\right)\)
\(m_{H_2SO_4\left(dư\right)}=m_{Al_2\left(SO_4\right)_3}=342.0,5\dfrac{161}{825}=\dfrac{9177}{275}\left(g\right)\)
=> \(m_{H_2SO_4\left(bđ\right)}=98\left(1,5a+b\right)+\dfrac{9177}{275}=\dfrac{22652}{275}\left(g\right)\)
=> \(y=\dfrac{\dfrac{22652}{275}.100}{10}=\dfrac{45304}{55}\left(g\right)\)
A) Gọi x,y tương ứng là số mol của Al và Fe:
Ta có: 27x+56y=11 (1)
\(n_{H_2}\)=0,4 mol
1,5x+y=0,4 (2)
Giải hệ(1),(2):x=0,2;y=0,1
\(m_{Al}=0,2.27=5,4g\)
%Al=\(^{\frac{5,4.100}{11}}=49,09\%\)
\(m_{Fe=0,1.56=5,6g}\)
%Fe=\(50,91\%\)
B)Nồng độ phần trăm của dd \(H_2SO_4\) là:
C%=\(\frac{m_{ct}}{m_{dd}}\times100\%\)=\(\frac{16,6}{200}\times100=8,3\%\)
a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) Gọi x,y là số mol Al, Fe
\(n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Ta có hệ : \(\left\{{}\begin{matrix}27x+56y=0,83\\\dfrac{3}{2}x+y=0,02\end{matrix}\right.\)
=> \(x=\dfrac{29}{5700};y=\dfrac{47}{3800}\)
\(\%m_{Al}=\dfrac{\dfrac{27}{5700}.27}{0,83}.100=16,55\%\); \(\%m_{Fe}=100-16,55=83,45\%\)
c)Bảo toàn nguyên tố H: \(n_{H_2SO_4}=n_{H_2}=0,02\left(mol\right)\)
=> \(C\%_{H_2SO_4}=\dfrac{0,02.98}{200}.100=0,98\%\)
Gọi x là số mol của Mg tham gia pứ
=> 0,5x là số mol Al tham gia pứ
nH2 = \(\dfrac{7,84}{22,4}=0,35\) mol
Pt: Mg + H2SO4 --> MgSO4 + H2
.....x............x.................x...........x
....2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
...0,5x.....0,75x............0,25x...........0,75x
Ta có: \(x+0,75x=0,35\)
Giải ra \(x=0,2\)
a) mMg tham gia pứ = 0,2 . 24 = 4,8 (g)
mAl tham gia pứ = \(\dfrac{0,2}{2}.27=2,7\left(g\right)\)
b) Theo pt ta có: nH2SO4 = nH2 = 0,35 mol
mH2SO4 = 0,35 . 98 = 34,3 (g)
C% dd H2SO4 = \(\dfrac{34,3}{343}.100\%=10\%\)
c) nMgSO4 = x = 0,2 mol
mMgSO4 = 0,2 . 120 = 24 (g)
nAl2(SO4)3 = 0,25x = 0,25 . 0,2 = 0,05 mol
mAl2(SO4)3 = 0,05 . 342 = 17,1 (g)
Áp dụng ĐLBTKL, ta có:
mdd sau pứ = mhh kim loại + mdd H2SO4 - mH2
....................= (4,8 + 2,7) + 343 - (0,35 . 2) = 349,8 (g)
C% dd MgSO4 = \(\dfrac{24}{349,8}.100\%=6,86\%\)
C% dd Al2(SO4)3 = \(\dfrac{17,1}{349,8}.100\%=5\%\)
thanks nha