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1.
nAl=\(\dfrac{5,4}{27}\)=0,2 mol
mHCl=\(\dfrac{175.14,6}{100}\)=25,55g
nHCl=\(\dfrac{25,55}{36,5}\)=0,7
2Al + 6HCl → 2AlCl3 + 3H2↑
n trước pứ 0,2 0,7
n pứ 0,2 →0,6 → 0,2 → 0,3 mol
n sau pứ hết dư 0,1
Sau pứ HCl dư.
mHCl (dư)= 36,5.0,1=3,65g
mcác chất sau pư= 5,4 +175 - 0,3.2= 179,8g
mAlCl3= 133,5.0,2=26,7g
C%ddHCl (dư)= \(\dfrac{3,65.100}{179,8}=2,03%\)%
C%ddAlCl3 = \(\dfrac{26,7.100}{179,8}\)= 14,85%
2.
200ml= 0,2l
mMg= \(\dfrac{4,2}{24}=0,175mol\)
Mg + 2HCl → MgCl2 + H2↑
0,175→ 0,35 → 0,175→0,175 mol
a) VH2= 0,175.22,4=3,92l.
b)C%dHCl= \(\dfrac{0,35}{0,2}=1,75\)M
\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(mHCl=\dfrac{200.7,3\%}{100\%}=14,6\left(g\right)\)
\(nHCl=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)
2 6 2 3 (mol)
0,1 0,3 0,1 0,15 (mol)
LTL : 0,1 / 2 < 0,4/6
=> Al đủ , HCl dư
1. \(VH_2=0,15.22,4=3,36\left(l\right)\)
2. \(mH_2=0,15.2=0,3\left(g\right)\)
mdd = mAl + mddHCl - mH2 = 2,7 + 200 - 0,3 = 202,4 (g)
\(mH_2SO_{4\left(dưsaupứ\right)}=0,1.98=9,8\left(g\right)\)
\(mAlCl_2=0,1.98=9,8\left(g\right)\)
\(C\%_{ddH_2SO_4}=\dfrac{9,8.100}{202,4}=4,84\%\)
\(C\%_{AlCl_2}=\dfrac{9,8.100}{202,4}=4,84\%\)
a)
$n_{HCl} = \dfrac{150.14,6\%}{36,5} = 0,6(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH :
$n_{H_2} = 0,3(mol)$
$n_{Al} = n_{AlCl_3} = \dfrac{1}{3}n_{HCl} = 0,2(mol)$
$m_{Al} = 0,2.27 = 5,4(gam)$
b)
$m_{dd} = 5,4 + 150 - 0,3.2 = 154,8(gam)$
$C\%_{AlCl_3} = \dfrac{0,2.133,5}{154,8}.100\% = 17,25\%$
nAl=0,2(mol)
mHCl=500.10%=50(g) => nHCl=50/36,5=100/73(mol)
PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
Vì: 0,2/2 < 100/73:6
=> Al hết, HCl dư, tính theo nAl
a) nH2=3/2. 0,2=0,3(mol) => V(H2,đktc)=0,3.22,4=6,72(l)
b) mHCl(tham gia p.ứ)= 6/2. 0,2 . 36,5= 21,9(g)
c) mddsau= 5,4+500-0,3.2=504,8(g)
mAlCl3=0,2. 133,5= 26,7(g)
mHCl(DƯ)= 50 -21,9=28,1(g)
C%ddAlCl3= (26,7/504,8).100=5,289%
C%ddHCl(dư)= (28,1/504,8).100=5,567%
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{400.14,6\%}{36,5}=1,6\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
Xét tỉ lệ: \(\dfrac{0,4}{2}< \dfrac{1,6}{6}\) => Al hết, HCl dư
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,4------------->0,4---->0,6
mdd sau pư = 10,8 + 400 - 0,6.2 = 409,6 (g)\(C\%_{AlCl_3}=\dfrac{0,4.133,5}{409,6}.100\%=13,037\%\)
\(m_{HCl}=150\cdot14.6=21.9\left(g\right)\)
\(n_{HCl}=\dfrac{21.9}{36.5}=0.6\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.2.........0.6..........0.2.......0.3\)
\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)
\(m_{dd}=5.4+150-0.3\cdot2=154.8\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{0.2\cdot133.5}{154.8}\cdot100\%=17.24\%\)
2Al+ 6HCl --------> 2AlCl3 + 3H2
\(n_{HCl}=\dfrac{150.14,6\%}{36,5}=0,6\left(mol\right)\)
Ta có : \(n_{Al}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\)
=> \(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{ddsaupu}=5,4+150-0,6.2=154,2\left(g\right)\)
=>\(C\%_{AlCl_3}=\dfrac{0,2.133,5}{154,2}.100=17,32\%\)
nNa = 6.9 : 23 = 0.3 mol
4Na + O2 ->2 Na2O
mol : 0.3 -> 0.15
Na2O + H2O -> 2NaOH
mol : 0.15 -> 0.3
mdd = 0.15 x 62 + 140.7 = 150g
C% NaOH = 0.3x40: 150 x 100% = 8%
a.b.\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,1 0,3 0,2 ( mol )
\(V_{H_2}=0,3.22,4=6,72l\)
c.\(n_{HCl}=\dfrac{125.14,6\%}{36,5}=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 < 0,5 ( mol )
0,2 0,2 0,2 ( mol )
\(m_{FeCl_2}=0,2.127=25,4g\)
\(m_{ddspứ}=\left(0,2.56\right)+125-0,2.2=135,8g\)
\(C\%_{FeCl_2}=\dfrac{25,4}{135,8}.100=18,7\%\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\:\right)\\
Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,1 0,3 0,2
=> \(m_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(m_{HCl}=125.14,6\%=18,25\left(g\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
Fe + 2HCl →FeCl2 +H2
\(C\%=\dfrac{11,2}{18,25}.100\%=61,3\%\)
2Al+6HCl->2AlCl3+3H2
0,185-----------0,185---0,2775 mol
n Al=\(\dfrac{5}{27}\)=0,185 mol
m HCl=21,9=>n HCl=0,6 mol
=>Al td hết , HCl dư
=>C% AlCl3=\(\dfrac{0,185.133,5}{5+150-0,2775.2}\).100=15,99%
=>C% HCl dư=\(\dfrac{0,045.36,5}{5+150-0,2775.2}\).100=1%