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\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\uparrow\)
0,1 → 0,1 → 0,1
a) \(V_{H_2}=22,4\cdot0,1=2,24\left(l\right)\)
b) \(m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\)
c) \(Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\uparrow\)
bđ: 0,1 → 0,4
pư: 0,1 → 0,1
\(\Rightarrow H_2SO_4\text{ dư}\)
\(\Rightarrow n_{H_2SO_4\text{ dư}}=0,4-0,1=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\text{ dư}}=0,3\cdot98=29,4\left(g\right)\)
\(nAl=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(nHCl=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
2 6 2 3 (mol)
0,2 0,6 0,2 0,3 (mol)
LTL : 0,3 / 2 > 0,6/6
=> Al dư sau pứ , HCl đủ vs pứ
\(mAl_{\left(dư\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
\(mAlCl_3=0,2.98=19,6\left(g\right)\)
\(H_2+CuO\rightarrow Cu+H_2O\)
1 1 1 1 (mol)
0,3 0,3 0,3 0,3 (mol)
=> \(mCu=0,3.64=19,2\left(g\right)\)
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\
n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\
pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(LTL:\dfrac{0,3}{2}>\dfrac{0,6}{6}\)
=> Al dư HCl hết
theo pthh : \(n_{Al\left(p\text{ư}\right)}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\\ m_{Al\left(d\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
theo pthh : \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,1\left(mol\right)\\
m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
theo pthh : \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)
pthh: \(CuO+H_2\underrightarrow{t^o}H_2O+Cu\)
0,3 0,3
\(m_{Cu}=0,3.64=19,2\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)
\(n_{HCl}=2,5.0,2=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 < 0,5 ( mol )
0,2 0,4 0,2 0,2 ( mol )
Chất dư là HCl
\(m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65g\)
\(V_{H_2}=0,2.22,4=4,48l\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1M\)
a) $Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : n H2 = n Fe = 8,4/56 = 0,15(mol)
V H2 = 0,15.22,4 = 3,36(lít)
b) n HCl = 2n Fe = 0,3(mol)
=> CM HCl = 0,3/0,2 = 1,5M
c) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy :
n CuO = 32/80 = 0,4 > n H2 = 0,15 mol nên CuO dư
Theo PTHH : n Cu = n H2 = 0,15 mol
=> m Cu = 0,15.64 = 9,6 gam
\(a,Zn+2HCl\to ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2(mol);n_{HCl}=\dfrac{21,9}{36,5}=0,6(mol)\)
Vì \(\dfrac{n_{Zn}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư
\(n_{HCl(dư)}=0,6-0,2.2=0,2(mol)\\ c,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2(mol)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2(g)\\ V_{H_2}=0,2.22,4=4,48(l)\)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\)
_________0,1__________0,1
\(b.n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
$n_{Zn} = \dfrac{6,5}{65} = 0,1(mol) \\ PTHH: Zn + 2HCl \to ZnCl_2 + H_2 \\$$n_{H_2} = n_{Zn} = 0,1(Mol) \\ V_{H_2} = 0,1.22,4 = 2,24l \\b) PTHH: H_2 + CuO \xrightarrow[]{t^o} Cu + H_2O \\ n_{CuO} = \dfrac{12}{64} = 0,15(mol) \\ \to CuO dư$ $\\ n_{H_2} = n_{Cu} = 0,1(mol \\ m_{Cu} = 0,1.64 = 6,4(gam)$
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
trc p/u : 0,1 0,4
p/u : 0,1 0,2 0,1 0,1
Sau : 0 0,2 0,1 0,1
a, ----> Sau p/u HCl dư
\(m_{HCldư}=0,2.26,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{HCl}=\dfrac{200.7,3}{100}=14,6\left(g\right)\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
b, \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
\(m_{ddZnCl_2}=6,5+200-\left(0,1.2\right)=206,3\left(g\right)\)
\(C\%_{ZnCl_2}=\dfrac{13,6}{206,3}.100\%\approx6,59\%\)
c, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
Thể tích H2 thu được thực tế :
\(V_{H_2tt}=2,24.90\%=2,016\left(l\right)\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2.......0.4......0.2...........0.2\)
\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(0.4............0.4\)
\(m_{NaOH\left(dư\right)}=\left(0.8-0.4\right)\cdot40=16\left(g\right)\)
\(a)\ n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{FeCl_2} = n_{Fe} = 0,2(mol)\\ \Rightarrow m_{FeCl_2} = 0,2.127 = 25,4\ gam\\ b) n_{H_2} = n_{Fe} = 0,2(mol)\Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)\\ c) n_{HCl} = 2n_{Fe} = 0,4(mol)\\ NaOH + HCl \to NaCl + H_2O\\ n_{NaOH} = 0,8> n_{HCl} = 4 \Rightarrow NaOH\ dư\\ n_{NaOH\ pư} = n_{HCl} = 0,4(mol)\\ \Rightarrow m_{NaOH\ dư} = (0,8-0,4).40 = 16\ gam\)