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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) 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: \(\left\{{}\begin{matrix}n_{CO_2}=n_{BaCl_2}=0,35\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,35\cdot22,4=7,84\left(l\right)\\C_{M_{BaCl_2}}=\dfrac{0,35}{0,25}=1,4\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\end{matrix}\right.\)
\(1,PTHH:4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ Al_2O_3+6HCl\to 2AlCl_3+3H_2\\ 2,n_{HCl}=\dfrac{240.7,3\%}{100\%.36,5}=0,48(mol)\\ \Rightarrow n_{Al_2O_3}=\dfrac{1}{6}n_{HCl}=0,08(mol)\\ \Rightarrow n_{Al}=2n_{Al_2O_3}=0,16(mol)\\ \Rightarrow m_{Al}=0,16.27=4,32(g)\\ n_{H_2}=\dfrac{1}{2}n_{HCl}=0,24(mol)\\ n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,16(mol)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,16.133,5}{0,08.102+240-0,24.2}.100\%=8,62\%\)
a, \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
0,2------------------>0,4
\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{12,4+50}.100\%=25,64\%\)
b, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
0,2------------------->0,2------->0,1
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40+16}{100+16+4,6-0,1.2}.100\%==20\%\)
c, \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\)
PTHH:
2Na + 2HCl ---> 2NaCl + H2
0,2<-----0,2-----------0,2--->0,1
2Na + 2H2O ---> 2NaOH + H2
0,2------------------>0,2----->0,1
\(\Rightarrow m_{dd}=9,2+100-\left(0,1+0,1\right).2=108,8\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,2.58,5}{108,8}.100\%=10,75\%\\C\%_{NaOH}=\dfrac{0,2.40}{108,8}.100\%=7,35\%\end{matrix}\right.\)
CaO + 2HCl \(\rightarrow\)CaCl2 + H2O
nCaO=\(\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PTHH ta có:
2nCaO=nHCl=0,4(mol)
mHCl=36,5.0,4=14,6(g)
m dd HCl=\(14,6:\dfrac{3,65}{100}=400\left(g\right)\)
b;
Theo PTHH ta có:
nCaO=nCaCl2=0,2(mol)
mCaCl2=111.0,2=22,2(g)
C% dd CaCl2=\(\dfrac{22,2}{11,2+400}.100\%=5,4\%\%\)
- Cảm ơn bạn, có thể giúp mình 1 số câu khác được không?
Bài 9:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{14,6\%.100}{36,5}=0,4\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,4-2.0,1=0,2\left(mol\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)
Em xem lại đề giúp anh nhé. Ban đầu là MgO, sao lúc sau lại là NaCl ?
a,
MgO + 2HCl → MgCl2 + H2O
(mol) \(\frac{a}{40}\)..............\(\frac{a}{20}\) ..............\(\frac{a}{40}\)
Ta có : \(\frac{a}{40}\) (24+ 35,5 *2) = a + 55
<=> a = 40 (gam)
=> nHCl = a / 20 = 40/20 = 2 (mol)
mà : nHCl = \(\frac{b\cdot3,65\%}{100\%\cdot36,5}\) = 2
=> b = 2000 (gam)
Dung dịch sau pư chỉ có MgCl2 .
m(dd sau) = m(Mg) + m(dd HCl) = a+ b = 40 + 2000 = 2040 (gam)
C%(MgCl2) = \(\frac{a+55}{2040}\cdot100\%\) = \(\frac{40+55}{2040}\cdot100\%\) ~ 4,65 (%)
\(_{ }\)nHCL =\(\dfrac{3,65}{100x36,5}\)x103=1 mol
nCAO=0,1 mol
CAO +2HCL→CACL2+ H2O
0,1---> 0,2
=> HCL DƯ