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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%
a)
\(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,03<------------0,03<----0,015
=> \(\%m_{Na}=\dfrac{0,03.23}{1,31}.100\%=52,67\%\)
=> \(\%m_{Na_2O}=100\%-52,67\%=47,33\%\)
b)
\(n_{Na_2O}=\dfrac{1,31.47,33\%}{62}=0,01\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
0,01----------->0,02
=> nNaOH = 0,03 + 0,02 = 0,05 (mol)
mdd sau pư = 1,31 + 18,72 - 0,015.2 = 20 (g)
=> \(C\%_{dd.NaOH}=\dfrac{0,05.40}{20}.100\%=10\%\)
\(V_{dd.NaOH}=\dfrac{20}{1,2}=\dfrac{50}{3}\left(ml\right)=\dfrac{1}{60}\left(l\right)\)
\(C_{M\left(dd.NaOH\right)}=\dfrac{0,05}{\dfrac{1}{60}}=3M\)
a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$
b) n Al = 8,1/27 = 0,3(mol)
Theo PTHH :
n H2 = 3/2 n Al = 0,45(mol)
V H2 = 0,45.22,4 = 10,08(lít)
c) n AlCl3 = n Al = 0,3(mol)
m AlCl3 = 0,3.133,5 = 40,05(gam)
d) n HCl = 3n Al = 0,9(mol)
m dd HCl = 0,9.36,5/7,3% = 450(gam)
Sau phản ứng :
m dd = 8,1 + 450 -0,45.2 = 457,2(gam)
C% AlCl3 = 40,05/457,2 .100% = 8,76%
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)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4----->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) mHCl = 0,4.36,5 = 14,6 (g)
=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)
c)
mdd sau pư = 13 + 200 - 0,2.2 = 212,6 (g)
mZnCl2 = 0,2.136 = 27,2 (g)
=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)
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%
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%
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%
\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\\ m_{H_2SO_4}=9,8\%.40=3,92\left(g\right)\\ n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\)
PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
LTL: \(\dfrac{0,02}{2}< \dfrac{0,04}{3}\rightarrow\)H2SO4 dư
Theo pt: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,02=0,03\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,02=0,01\left(mol\right)\end{matrix}\right.\)
\(\rightarrow V_{H_2}=0,03.22,4=0,672\left(l\right)\\ m_{dd}=0,54+40=40,54\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,01}{40,54}=8,43\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,04-0,03\right).98}{40,54}=2,41\%\end{matrix}\right.\)