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a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,35\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,35.65=22,75\left(g\right)\)
b, Theo PT: \(n_{HCl\left(pư\right)}=2n_{H_2}=0,7\left(mol\right)\)
Mà: axit dùng dư 10% so với lượng pư.
\(\Rightarrow n_{HCl}=0,7+0,7.10\%=0,77\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,77}{2,8}=0,275\left(l\right)\)
c, Ta có: \(D=\dfrac{m}{V}\Rightarrow m_{ddHCl}=0,275.1000.1,04=286\left(g\right)\)
d, Theo PT: \(n_{ZnCl_2}=n_{H_2}=0,35\left(mol\right)\)
Dd X gồm: ZnCl2 và HCl dư.
nHCl dư = 07.10% = 0,07 (mol)
Ta có: m dd sau pư = mZn + m dd HCl - mH2 = 308,05 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,35.136}{308,05}.100\%\approx15,452\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,07.36,5}{308,05}.100\%\approx0,829\%\end{matrix}\right.\)
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)
=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)
`a)PTHH:`
`Zn + 2HCl -> ZnCl_2 + H_2`
`0,1` `0,2` `0,1` `0,1` `(mol)`
`n_[HCl]=0,2.1=0,2(mol)`
`=>m_[Zn]=0,1.65=6,5(g)`
`b)m_[dd HCl]=1,1.200=220(g)`
`=>C%_[ZnCl_2]=[0,1.136]/[6,5+220-0,1.2].100~~6%`
\(a,n_{HCl}=0,2.1=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1<--0,2------>0,1------->0,1
\(\rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
\(b,m_{ddHCl}=200.1,1=220\left(g\right)\)
\(\rightarrow m_{dd}=220+6,5-0,1.2=226,3\left(g\right)\\ \rightarrow C\%_{ZnCl_2}=\dfrac{0,1.136}{226,3}.100\%=6\%\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,4 0,8 0,4 0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)
a) \(PT:CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
\(HCl+NaOH\rightarrow NaOH+H_2O\)
b) \(m_{HCl}=\frac{200.10,95\%}{100\%}=21,9\left(g\right)\)
\(n_{HCl}=\frac{21,9}{36,5}=0,6\left(mol\right)\)
c) \(n_{NaOH}=2.0,05=0,1\left(mol\right)\Rightarrow n_{HCl\left(pưNaOH\right)}=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl\left(pưCaCO_3\right)}=0,6-0,1=0,5\left(mol\right)\)
d) \(n_{CaCO_3}=\frac{1}{2}n_{HCl\left(pưCaCO_3\right)}=0,5.\frac{1}{2}=0,25\left(mol\right)\)
\(m_{CaCO_3}=0,25.100=25\left(g\right)\)
e) \(n_{CO_2}=n_{CaCO_3}=0,25\left(mol\right)\)
\(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
f) \(n_{CaCl_2}=n_{CaCO_3}=0,25\left(mol\right)\)
\(m_{ddA}=25+200-0,25.44=214\left(g\right)\)
\(C\%_{ddCaCl_2}=\frac{0,25.111}{214}.100\%=12,97\%\)
\(C\%_{ddHCldư}=\frac{0,1.36,5}{214}.100\%=1,71\%\)
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