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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)
=> \(C\%_{dd.NaOH}=\dfrac{0,05.40}{20}.100\%=10\%\)
\(V_{dd.NaOH}=\dfrac{20}{1,2}=\dfrac{50}{3}\left(ml\right)=\dfrac{1}{60}\left(l\right)\)
\(C_{M\left(dd.NaOH\right)}=\dfrac{0,05}{\dfrac{1}{60}}=3M\)
nP2O5 = 14,2/142 = 0,1 (mol)
PTHH: P2O5 + 3H2O -> 2H3PO4
Mol: 0,1 ---> 0,3 ---> 0,2
CMddH3PO4 = 0,2/0,5 = 0,4M
Chất rắn D là Cu, chất rắn E là CuO
\(m_{tăng}=m_{O_2}=0,16\left(g\right)\)
=> \(n_{O_2}=\dfrac{0,16}{32}=0,005\left(mol\right)\)
PTHH: 2Cu + O2 --to--> 2CuO
0,01<-0,005
=> mCu = 0,01.64 = 0,64 (g)
Gọi số mol K, Ba là a, b (mol)
=> 39a + 137b = 3,18 - 0,64 = 2,54 (1)
PTHH: 2K + 2H2O --> 2KOH + H2
a--------------->a
Ba + 2H2O --> Ba(OH)2 + H2
b--------------->b
=> 56a + 171b = 3,39 (2)
(1)(2) => a = 0,03 (mol); b = 0,01 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,64}{3,18}.100\%=20,126\%\\\%m_K=\dfrac{0,03.,39}{3,18}.100\%=36,792\%\\\%m_{Ba}=\dfrac{0,01.137}{3,18}.100\%=43,082\%\end{matrix}\right.\)
\(m_{O_2}=m+0,16-m=0,16\left(g\right)\\ \rightarrow n_{O_2}=\dfrac{0,16}{32}=0,005\left(mol\right)\)
PTHH: 2Cu + O2 --to--> 2CuO
0,01 0,005
Gọi \(\left\{{}\begin{matrix}n_K=a\left(mol\right)\\n_{Ba}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
2K + 2H2O ---> 2KOH + H2
a a
Ba + 2H2O ---> Ba(OH)2 + H2
b b
Hệ pt \(\left\{{}\begin{matrix}39a+137b=3,18-0,01.64=2,54\\56a+171b=3,39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,03\left(mol\right)\\b=0,01\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,01.64}{3,18}=20,13\%\\\%m_K=\dfrac{0,03.39}{3,18}=36,79\%\\\%m_{Ba}=100\%-20,13\%-36,79\%=43,08\%\end{matrix}\right.\)
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) => \(\left\{{}\begin{matrix}a=\dfrac{161}{825}\left(mol\right)\\b=\dfrac{57}{275}\left(mol\right)\end{matrix}\right.\)
=> \(x=\dfrac{161}{825}.27+\dfrac{57}{275}.65=\dfrac{5154}{275}\left(g\right)\)
\(m_{H_2SO_4\left(dư\right)}=m_{Al_2\left(SO_4\right)_3}=342.0,5\dfrac{161}{825}=\dfrac{9177}{275}\left(g\right)\)
=> \(m_{H_2SO_4\left(bđ\right)}=98\left(1,5a+b\right)+\dfrac{9177}{275}=\dfrac{22652}{275}\left(g\right)\)
=> \(y=\dfrac{\dfrac{22652}{275}.100}{10}=\dfrac{45304}{55}\left(g\right)\)
\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\\Tacó:n_{BaO} =n_{Ba\left(OH\right)_2}=0,25.0,4=0,1\left(mol\right)\\ \Rightarrow m_{BaO}=0,1.153=15,3\left(g\right)\\ \Rightarrow\%m_{CuO}=\dfrac{20-15,3}{20}.100=30,72\%\)
1. `2Na + 2H_2O -> 2NaOH + H_2`
2. chất rắn không tan là Cu `=> m_{Cu} = 7,7 (g)`
`=> m_{Na} = 10 - 7,7 = 2,3 (g) => n_{Na} = (2,3)/(23) = 0,1 (mol)`
`n_{H_2} = 1/2 n_{Na} = 0,05 (mol)`
`=> V = V_{H_2} = 0,05.22,4 = 1,12 (l)`
3. `n_{NaOH} = n_{Na} = 0,1 (mol)`
`=> C_{M(NaOH)} = (0,1)/(0,2) = 0,5M`