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n Zn= 19,5/65=0,3 (mol).
PTPƯ: Zn(0.3) + HCl(0.6) ----> ZnCl2(0.3) + H2(0,3)
mHCl=0,6.36.5=21.9(g)
a) C%HCl= 21.9/300.100%=7,3%
b) VH2=0,3.22,4=6,72(lít)
c) mH2=0,3.2=0,6(g)
mZnCl2=0,3.136=40,8(g)
mddZnCl2 =(19,5+300)-0,6=318,9(g)
C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%
Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,1-->0,2------------------>0,1
=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)
a, \(m_{HCl}=150.7,3\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{Mg}=n_{MgCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(m_{Mg}=0,15.24=3,6\left(g\right)\)
b, \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
c, Ta có: m dd sau pư = 3,6 + 150 - 0,15.2 = 153,3 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,15.95}{153,3}.100\%\approx9,3\%\)
a)
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(m_{HCl}=\dfrac{175.14,6}{100}=25,55\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{25,55}{35,5}=0,7\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
bđ 0,3 0,7
pư 0,3 0,6
spư 0 0,1 0,3 0,3
=> VH2 = 0,3.22,4 = 6,72 (l)
b)
mdd = 16,8 + 175 - 0,3.2 = 191,2 (g)
=> \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,93\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{191,2}.100\%=1,91\%\end{matrix}\right.\)
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{\dfrac{175.14,6}{100}}{36,5}=0,7\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(LTL:\dfrac{0,3}{1}< \dfrac{0,7}{2}\)
\(n_{H_2}=n_{Fe}=0,3\left(mol\right)\\
V_{H_2}=0,3.22,4=6,72\left(l\right)\\
m_{\text{dd}}=16,8+175-\left(0,3.2\right)=191,2\left(g\right)\\
n_{FeCl_2}=n_{Fe}=0,3\left(mol\right)\\
C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,92\%\)
=> HCl dư
Sửa đề: 8,4 gam Fe
\(a,n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ n_{HCl}=\dfrac{14,6.175}{36,5.100}=0,7\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
ban đầu 0,15 0,7
phản ứng 0,15 0,3
sau pư 0 0,4 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
\(b,m_{dd}=8,4+175-0,15.2=183,1\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{183,1}.100\%=10,4\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,4.36,5}{183,1}.100\%=7,97\%\end{matrix}\right.\)
a) \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,05->0,1----->0,05---->0,05
`=> V_{ddHCl} = (0,1)/2 = 0,05 (l)`
b) `V_{H_2} = 0,05.22,4 = 1,12 (l)`
c) `C_{M(FeCl_2)} = (0,05)/(0,05) = 1M`
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\) ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
Ta có: m dd sau pư = 6,5 + 100 - 0,1.2 = 106,3 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b. số mol của 16,8 gam Fe:
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
Khối lượng của HCl:
\(m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)
c.Thể tích khí Hiđro (đktc):
\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)
\(n_K=\dfrac{9,75}{39}=0,25\left(mol\right)\)
\(n_{HCl}=\dfrac{300.7,3\%}{36,5}=0,6\left(mol\right)\)
PTHH: 2K + 2HCl --> 2KCl + H2
Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,6}{2}\) => HCl dư
PTHH: 2K + 2HCl --> 2KCl + H2
0,25-->0,25-->0,25-->0,125
=> VH2 = 0,125.22,4 = 2,8 (l)
mKCl = 0,25.74,5 = 18,625 (g)
mdd sau pư = 9,75 + 300 - 0,125.2 = 309,5 (g)
mHCl(dư) = (0,6 - 0,25).36,5 = 12,775 (g)
\(\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{18,625}{309,5}.100\%=6,018\%\\C\%_{HCl}=\dfrac{12,775}{309,5}.100\%=4,128\%\end{matrix}\right.\)