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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
Sửa lại câu c .
\(n_{H_2SO_4}=\dfrac{49.40}{100}:98=0,2\left(mol\right)\)
\(PTHH:\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
trc p/u : 0,3 0,2
p/u : 0,2 0,2 0,2 0,2
sau : 0,1 0 0,2 0,2
-> Fe dư
\(m_{ddFeSO_4}=0,3.56+49-0,4=65,4\left(g\right)\) ( ĐLBTKL )
\(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
\(C\%=\dfrac{30,4}{65,4}.100\%\approx46,48\%\)
PTHH :
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,3 0,3 0,3 0,3
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(a,m_{Fe}=0,3.56=16,8\left(g\right)\)
\(b,C_M=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)
\(c,n_{H_2SO_4}=\dfrac{\dfrac{49.40}{100}}{98}=0,2\left(mol\right)\)
\(\rightarrow n_{FeSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
\(m_{ddFeSO_4}=49+\left(0,2.56\right)-0,2.2=59,8\left(g\right)\)( định luật bảo toàn khối lượng )
\(C\%=\dfrac{30,4}{59,8}.100\%\approx50,84\%\)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4-->0,6---------->0,2------->0,6
=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)
b) VH2 = 0,6.22,4 = 13,44 (l)
c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\
pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,4 0,6 0,2 0,6
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\)
\(C_M=\dfrac{0,2}{0,15}=1,3M\)
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)
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(m_{HCl}=\dfrac{175.14,6}{100}=25,55\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{25,55}{35,5}=0,7\left(mol\right)\)
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
1)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15->0,3--->0,15-->0,15
=> VH2 = 0,15.22,4 = 3,36 (l)
c) mdd sau pư = 8,4 + 250 - 0,15.2 = 258,1 (g)
=> \(C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%=7,38\%\)
2)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4---->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
mZnCl2 = 0,2.136 = 27,2 (g)
c) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,2}=2M\)
d)
PTHH: A + 2HCl --> ACl2 + H2
0,2<--0,4
=> \(M_A=\dfrac{4,8}{0,2}=24\left(g/mol\right)\)
=> A là Mg(Magie)
\(n_{Ba}=\dfrac{6,85}{137}=0,05\left(mol\right)\\ m_{H_2SO_4}=500.1,96\%=9,8\left(g\right)\\ PTHH:Ba+H_2SO_4\rightarrow BaSO_4+H_2\uparrow\\ LTL:0,05< 0,1\Rightarrow H_2SO_4.dư\)
\(n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{Ba}=n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(V_{dd}=\dfrac{500}{1,15}\approx434\left(ml\right)=0,434\left(l\right)\)
\(C_{MBaSO_4}=\dfrac{0,05}{0,434}=0,115M\\ C_{MH_2SO_4\left(dư\right)}=\dfrac{0,05}{0,434}=0,115M\)
nMg = mMg / M(Mg) = 3 / 24 = 0,125 (mol)
m\(_{H2SO4}\) = (m\(_{dd}\) * C%) / 100% = ( 150 * 1,96%) / 100% = 2,94 (gam)
nH2SO4 = m\(_{H2SO4}\) / M\(_{H2SO4}\) = 2,94 / 98 = 0,03 (mol)
a, Mg + H2SO4 → MgSO4 + H2
(mol) 0,125......0,03
xét tỉ lệ : \(\frac{0,125}{1}\) > \(\frac{0,03}{1}\)
pư : 0,03.................0,03...........0,03...........0,03
dư : (0,125-0,03)
0,095 (mol)
vậy : V(H2) = 0,03 * 22,4 = 0,672 (lít)
b, Dung dịch sau pư có MgSO4 : 0,03 (mol)
mặt khác Mg còn dư sau pư : 0,095 mol
m(dd sau) = m(Mg bđ) + m(dd H2SO4 ) - m(H2) - m(Mg dư)
= 3 + 150 - 0,03 * 2 - 0,095*24
= 150,66 (gam)
Vậy : C%(MgSO4) = {mMgSO4 / m(dd sau) } * 100%
= (0,03*120/150,66) * 100%
~ 2,39 (%)
nMg= 3/24=0.125 mol
mH2SO4= 150*1.96/100=2.94g
nH2SO4= 2.94/98=0.03 mol
Mg + H2SO4 --> MgSO4 + H2
Bđ: 0.125__0.03
Pư : 0.03___0.03______0.03____0.03
Kt: 0.095___0_________0.03___0.03
VH2= 0.03*22.4=0.672l
mdd sau phản ứng = mMg(bđ) + mddH2SO4 - mH2 - mMg(dư)
=> 3 + 150 - 0.06 - 0.095*24=150.66g
mMgSO4= 0.03*120=3.6g
C%MgSO4= 3.6/150.66*100%= 2.39%