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\(a) n_{Zn} = \dfrac{19,5}{65} = 0,3(mol) ; n_{HCl} = 0,35.2 = 0,7(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{HCl} = 0,7 > 2n_{Zn} = 0,6 \to HCl\ dư\\ n_{H_2} = n_{Zn} = 0,3(mol) \Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{dd\ HCl} = 350.1,05 = 367,5(gam)\\ m_{dd\ sau\ pư} = 19,5 + 367,5 - 0,3.2 = 386,4(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,3.136}{386,4}.100\% = 10,56\%\\ c) C\%_{HCl} = \dfrac{0,7.36,5}{367,5}.100\% = 6,95\%\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,4 0,8 0,4 0,4
a) \(n_{H2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,4.22,4=8,96\left(l\right)\)
b) \(n_{HCl}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
⇒ \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)
\(m_{ddHCl}=\dfrac{29,2.100}{14,6}=200\left(g\right)\)
c) \(n_{ZnCl2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,4.136=54,4\left(g\right)\)
\(m_{ddspu}=26+200-\left(0,4.2\right)=225,2\left(g\right)\)
\(C_{ZnCl2}=\dfrac{54,4.100}{225,2}=24,16\)0/0
Chúc bạn học tốt
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
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) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{HCl} = 3n_{Al} = 0,6(mol)$
$m_{HCl} = 0,6.36,5 = 21,9(gam)$
b) $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$
c)
$m_{dd\ sau\ pư} = 5,4 + 200 - 0,3.2 = 204,8(gam)$
$m_{HCl\ dư} = 200.20\% - 21,9 = 18,1(gam)$
$C\%_{HCl} = \dfrac{18,1}{204,8}.100\% = 8,84\%$
$C\%_{AlCl_3} = \dfrac{0,2.133,5}{204,8}.100\% = 13,04\%$
a) nAl=5,427=0,2(mol)nAl=5,427=0,2(mol)
2Al+6HCl→2AlCl3+3H22Al+6HCl→2AlCl3+3H2
nHCl=3nAl=0,6(mol)
mHCl=0,6.36,5=21,9(gam)mHCl=0,6.36,5=21,9(gam)
b) nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)
VH2=0,3.22,4=6,72(lít)VH2=0,3.22,4=6,72(lít)
c)
mdd sau pư=5,4+200−0,3.2=204,8(gam)
mHCl dư=200.20%−21,9=18,1(gam)mHCl dư=200.20%−21,9=18,1(gam)
C%HCl=18,1204,8.100%=8,84%C%HCl=18,1204,8.100%=8,84%
C%AlCl3=0,2.133,5204,8.100%=13,04%
\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m=m_{Zn}=0,1.65=6,5\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ d,m_{ddZnCl_2}=6,5+100-0,1.2=106,3\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{106,3}.100\approx12,794\%\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2...................0.2..........0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)
PTHH : \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
a)Số mol của \(Al_2O_3\)là :
\(n_{Al_2O_3}=\frac{m_{Al_2O_3}}{M_{Al_2O_3}}=\frac{10,2}{102}=0,1\left(mol\right)\)
Theo PTHH ,ta có : \(n_{HCl}=n_{Al_2O_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl}=n_{HCl}.M_{HCl}=0,1.36,5=3,65\left(g\right)\)
b)Theo PTHH ,ta có : \(n_{HCl}=n_{AlCl_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
\(\Rightarrow C\%_{AlCl_3}=\frac{mAl_2O_3}{m_{AlCl_3}}=\frac{10,2}{13,35}\approx76,4\%\)
Ta có \(n_{Al_2O_3}=\frac{m}{M}=\frac{10,2}{102}=0,1\)(mol) (1)
Phương trinh hóa học phản ứng
Al2O3 + 6HCl ---> 2AlCl3 + 3H2O
1 : 6 : 2 : 3 (2)
Từ (1) và (2) => nHCl = 0,6 mol
=> mHCl = \(n.M=0,6.36,5=21,9\left(g\right)\)
Ta có \(\frac{m_{HCl}}{m_{dd}}=20\%\)
<=> \(\frac{21,9}{m_{dd}}=\frac{1}{5}\)
<=> \(m_{dd}=109,5\left(g\right)\)
=> Khối lượng dung dịch HCl 20% là 109,5 g
b) \(n_{AlCl_3}=0,2\)(mol)
=> \(m_{AlCl_3}=n.M=0,2.133,5=26,7g\)
mdung dịch sau phản ứng = 109,5 + 10,2 = 119,7 g
=> \(C\%=\frac{26,7}{119,7}.100\%=22,3\%\)
nzn= 6,5/65 =0.1mol
mHCl =50.18,25/100 =9,1g
=>nHCl= 9,1/36,5 =0,25mol
Zn + 2HCl -> ZnCl2 + H2
0,1 -> 0,2 -> 0,1 -> 0.1
0.1/1<0,25/2
=>nHCldư.Tính nzn
a) VH2= 0,1.22,4= 2,24l
b) mZnCl2= 0.1.(65+35,5.2)= 13,6g
c) mH2= 0,1.2 =0,2g
mdds= mzn+mddHCl-mH2= 56,3g
C%ZnCl2= 13,6.100/56,3 =24,2%
nHCldư= 0,25-0,2 =0,05mol
mHCldư= 0,05.(1+35,5) =1,825g
C%HCldư= 1,825.100/56,3 =3,2%
a)
Zn + 2HCl → ZnCl2 + H2
nHCl = 50.7,3% = 3,65 gam
<=> nHCl = 3,65 : 36,5 = 0,1 mol
Theo tỉ lệ phản ứng => nZn = 1/2nHCl = 0,05 mol
<=> mZn = 0,05.65 = 3,25 gam.
b) nH2 = 1/2nHCl = 0,05 mol
=> VH2 = 0,05 . 22,4 = 1,12 lít.
c) m dung dịch sau phản ứng = mZn + m dd HCl - mH2 = 3,25 + 50 - 0,05.2 = 53,15 gam
mZnCl2 = 0,05.136 = 6,8 gam
<=> C% ZnCl2 = \(\dfrac{6,8}{53,15}.100\%\) = 12,8%
\(n_{Zn}=\dfrac{19.5}{65}=0.3\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.3........0.6.........0.3......0.3\)
\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)
\(C\%_{HCl}=\dfrac{10.95}{200}\cdot100\%=5.475\%\)
\(m_{\text{dung dịch sau phản ứng}}=19.5+200-0.3\cdot2=218.9\left(g\right)\)
\(m_{ZnCl_2}=0.3\cdot136=40.8\left(g\right)\)
\(C\%_{ZnCl_2}=\dfrac{40.8}{218.9}\cdot100\%=18.63\%\)