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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
\(a.n_{H_2SO_4}=\dfrac{294.10\%}{98}=0,3\left(mol\right)\\ n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\\ MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\\ \rightarrow H_2SO_4dư\\ n_{MgSO_4}=n_{H_2SO_4\left(p.ứ\right)}=n_{MgO}=0,2\left(mol\right)\\ n_{H_2SO_4\left(p.ứ\right)}=0,3-0,2=0,1\left(mol\right)\\ m_{ddsau}=8+294=302\left(g\right)\\ b.C\%_{ddH_2SO_4\left(Dư\right)}=\dfrac{0,1.98}{302}.100\approx3,245\%\\ C\%_{ddMgSO_4}=\dfrac{0,2.120}{302}.100\approx7,947\%\)
Ta có:
n MgO = 0,2 ( mol )
m H2SO4 = 294 . 10% = 29,4 ( g )
=> n H2SO4 = 0,3 ( mol )
PTHH
MgO + H2SO4 ====> MgSO4 + H2O
0,2-------0,2-----------------0,2
theo pthh: n H2SO4 phản ứng = n Mg = 0,2 ( mol )
=> n H2SO4 dư = 0,1 ( mol )
BTKL:
m dd sau phản ứng = 8 + 294 = 302 ( g )
=> %m H2SO4 dư = 3,25 %
%m MgSO4 = 7,95%
a) PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{BaCO_3}=\dfrac{68,95}{197}=0,35\left(mol\right)\\n_{HCl}=0,25\cdot3,2=0,8\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,35}{1}< \dfrac{0,8}{2}\) \(\Rightarrow\) Axit còn dư
\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,35\cdot2=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)
b+c) Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{BaCl_2}=0,35\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,35\cdot22,4=7,84\left(l\right)\\C_{M_{BaCl_2}}=\dfrac{0,35}{0,25}=1,4\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\end{matrix}\right.\)
a)b)c)d) mBaCl2=150.16,64%=24,96g
=>nBaCl2=0,12 mol
mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol
BaCl2 + H2SO4 =>BaSO4 +2HCl
Bđ: 0,12 mol; 0,15 mol
Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol
Dư: 0,03 mol
Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol
Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol
mHCl=0,24.36,5=8,76g
mH2SO4=0,03.98=2,94g
Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g
mddA=mddBaCl2+mddH2SO4-mBaSO4
=150+100-27,96=222,04g
C%dd HCl=8,76/222,04.100%=3,945%
C% dd H2SO4=2,94/222,04.100%=1,324%
e) HCl +NaOH =>NaCl +H2O
0,24 mol=>0,24 mol
H2SO4 +2NaOH =>Na2SO4 + 2H2O
0,03 mol=>0,06 mol
TÔNG nNaOH=0,3 mol
=>V dd NaOH=0,3/2=0,15 lit
\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\\a, ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=98.0,3=29,4\left(g\right)\\ c,n_{ZnSO_4}=0,1.161=16,1\left(g\right)\\ m_{ddsau}=m_{ZnO}+m_{ddH_2SO_4}=8,1+200=208,1\left(g\right)\\ \Rightarrow C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{29,4}{208,1}.100\approx14,128\%\\ C\%_{ddZnSO_4}=\dfrac{16,1}{208,1}.100\approx7,737\%\)
ZnO+H2SO4->ZnSO4+H2O
0,1-----0,1-------0,1-------0,1 mol
n ZnO=\(\dfrac{8,1}{81}\)=0,1 mol
m H2SO4 =39,2g =>n H2SO4=\(\dfrac{39,2}{98}\)=0,4 mol
=>H2SO4 , dư 0,3 mol
=>m H2SO4=0,3.98=29,4g
=>C%H2SO4 dư=\(\dfrac{29,4}{200+0,1.18}\).100=14,568%
=>C% ZnSO4=\(\dfrac{0,1.161}{200+0,1.18}.100=7,9781\%\)
a, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 -----> FeSO4 + H2
b, ko tính đc do thiếu khối lượng ddH2SO4
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
nFe = 5,6 : 56 = 0,1 (mol)
nH2SO4 (đủ) = 0,1 (mol)
mH2SO4 = 0,1 . 98 = 9,8 (g)
\(mH_2SO_{\text{4(thamgiapứ) }}=\dfrac{9,8.100}{49}=20\left(g\right)\)
H2SO4 dư , Fe đủ
mH2SO4 dư = 20 - 9,8 = 10,2(g)
mFeSO4 = 0,1 . 152 = 15,2(g)
VH2 = 0,1 .22,4 = 2,24(l)
mH2 = 0,1 . 2 = 0,2 (g)
\(C\%H_2SO_4=\dfrac{10,2.100}{5,6+20+15,2-0,2}=25\%\)
\(C\%_{FeSO_4}=\dfrac{15,2.100}{5,6+20+15,2-0,2}=37\%\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot4,9\%}{98}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Cả 2 chất p/ứ hết
b+c) Theo PTHH: \(n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnSO_4}=0,1\cdot161=16,1\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}=206,3\left(g\right)\)
\(\Rightarrow C\%_{ZnSO_4}=\dfrac{16,1}{206,3}\cdot100\%\approx7,8\%\)
\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2......0.4..........0.2........0.2\)
\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)
\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)
\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)
\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)
a)
$CuO + H_2SO_4 \to CuSO_4 + H_2O$
b)
n CuO = 2,8/80 = 0,035(mol)
n H2SO4 = 140.20%/98 = 2/7 (mol)
Ta thấy :
n CuO / 1 < n H2SO4 /1 nên H2SO4 dư
n H2SO4 pư = n CuO = 0,035(mol)
=> m H2SO4 dư = 140.20% - 0,035.98 = 24,57(gam)
c)
m dd = m CuO + m dd H2SO4 = 2,8 + 140 = 142,8(gam)
C% CuSO4 = 0,035.160/142,8 .100% = 3,92%
C% H2SO4 = 24,57/142,8 .100% = 17,2%
Cảm ơn, cảm ơn bạn rất nhiều