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\(n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)
Gọi \(n_{NaOH}=a\left(mol\right)\rightarrow n_{NaCl}=1,5a\left(mol\right)\)
PTHH:
2Na + 2H2O ---> 2NaOH + H2
a a 0,5a
2Na + 2HCl ---> 2NaCl + H2
1,5a 1,5a 1,5 0,075a
\(\rightarrow0,5a+0,075a=0,025\\ \rightarrow a=0,2\left(mol\right)\)
\(\rightarrow n_{Na}=0,2+0,2.1,5=0,5\left(mol\right)\\ m_{Na}=0,5.23=11,5\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{100}=7,3\%\\C_{M\left(HCl\right)}=\dfrac{0,2}{\dfrac{100}{1,1}}=0,002M\end{matrix}\right.\)
a) $Fe + 2HCl \to FeCl_2 + H_2$
b)
n Fe = 8,4/56 = 0,15(mol) ; n HCl = 0,15.2,4 = 0,36(mol)
Ta thấy :
n Fe / 1 < n HCl /2 nên HCl dư
Theo PTHH : n H2 = n Fe = 0,15 mol
=> V = 0,15.22,4 = 3,36 lít
c) Dung dịch chứa HCl,FeCl2
m dd HCl = D.V = 0,8.150 = 120(gam)
Sau phản ứng :
n HCl dư = 0,36 - 0,15.2 = 0,06(mol)
n FeCl2 = n Fe = 0,15(mol)
m dd = 8,4 + 120 -0,15.2 = 128,1(gam)
C% HCl = 0,06.36,5/128,1 .100% = 1,71%
C% FeCl2 = 0,15.127/128,1 .100% = 14,87%
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{HCl}=0,15.4=0,6\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Zn}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)
c, \(n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow C_{M_{ZnCl_2}}=\dfrac{0,3}{0,15}=2\left(M\right)\)
a)Quy \(\left\{{}\begin{matrix}Na:x\left(mol\right)\\Ba:y\left(môl\right)\\O:z\left(mol\right)\end{matrix}\right.\underrightarrow{+H_2O}\left\{{}\begin{matrix}NaOH:x\left(mol\right)\\Ba\left(OH\right)_2:y\left(mol\right)\\O^{2-}:z\left(mol\right)\end{matrix}\right.+H_2\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)
\(n_{Ba\left(OH\right)_2}=\dfrac{20,52}{171}=0,12mol\Rightarrow y=0,12mol\)
Ta có hệ: \(\left\{{}\begin{matrix}BTKL:23x+137y+16z=21,9\\y=0,12\\BTe:x+2y=2z+2n_{H_2}\Rightarrow x-2z=-0,14\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,14\\y=0,12\\z=0,14\end{matrix}\right.\)
\(n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,14+2\cdot0,12=0,38mol\)
\(n_{CO _2}=\dfrac{6,72}{22,4}=0,3mol\Rightarrow n_{CO_3^{2-}}=0,38-0,3=0,08mol\)
\(\Rightarrow m_{CO_3^{2-}\downarrow}=0,08\cdot197=15,76g\)
1.
Al2O3 + 2NaOH -> 2NaAlO2 + H2O (1)
nNaAlO2=0,225(mol)
Từ 1:
nNaOH=nNaAlO2=0,225(mol)
nal2O3=\(\dfrac{1}{2}\)nNaAlO2=0,1125(mol)
V dd NaOH=0,225:5=0,045(lít)
mAl2O3=0,1125.102=11,475(g)
mquặng=11,475.110%=12,6225(g)
