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Lập PTHH theo các sơ đồ sau
CaCO3+2HCl-->CaCl2+H2O+CO2
BaCO3+2HCl-->BaCl2+CO2+H2O
K2CO3+2HBr-->2KBr+CO2+H2O
Na2CO3+2HCl-->2NaCl+CO2+H2O
MgCO3+2HNO3-->Mg(NO3)2+CO2+H2O
NaHCO3+HCl-->NaCl+CO2+H2O
NaHCO3+HNO3-->NaNO3+CO2+H2O
2NaHCO3+H2SO4-->Na2SO4+2CO2+2H2O
2Na2CO3+HNO3-->2NaNO3+CO2+H2O
CuCO3+2HCl-->CuCl2+CO2+H2O
CaCO3+2HCl-->CaCl2+H2O+CO2
BaCO3+2HCl-->BaCl2+CO2+H2O
K2CO3+2HBr-->2KBr+CO2+H2O
Na2CO3+2HCl-->2NaCl+CO2+H2O
MgCO3+2HNO3-->Mg(NO3)2+CO2+H2O
NaHCO3+HCl-->NaCl+CO2+H2O
NaHCO3+HNO3-->NaNO3+CO2+H2O
2NaHCO3+H2SO4-->Na2SO4+2CO2+2H2O
Na2CO3+2HNO3-->2NaNO3+CO2+H2O
CuCO3+2HCl-->CuCl2+CO2+H2O
a)
$n_{HCl} = \dfrac{250.14,6\%}{36,5} = 1(mol)$
$Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$
$n_{CO_2} = \dfrac{1}{2}n_{HCl} = 0,5(mol)$
$V_{CO_2} = 0,5.22,4 = 11,2(lít)$
b)
Sau phản ứng :
$m_{dd} = 55 + 250 -0,5.44 = 283(gam)$
$n_{Na_2CO_3} = n_{CO_2} = 0,5(mol) \Rightarrow m_{Na_2SO_4} = 55 - 0,5.106 = 2(gam)$
$n_{NaCl} =n_{HCl} = 1(mol)$
$C\%_{NaCl} = \dfrac{1.58,5}{283}.100\% = 20,67\%$
$C\%_{Na_2SO_4} = \dfrac{2}{283}.100\% = 0,71\%$
a) \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
\(n_{HCl}=\dfrac{250.14,6\%}{36,5}=1\left(mol\right)\)
\(TheoPT:n_{CO_2}=\dfrac{1}{2}n_{HCl}=0,5\left(mol\right)\)
=> \(V_{CO_2}=0,5.22,4=11,2\left(l\right)\)
b) \(C\%_{NaCl}=\dfrac{0,5.58,5}{55+250-0,5.44}.100=10,34\%\)
\(m_{Na_2SO_4}=55-0,5.106=2\left(g\right)\)
=> \(C\%_{Na_2SO_4}=\dfrac{2}{55+250-0,5.44}.100=0,7\%\)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ 0,1mol:0,2mol\rightarrow0,2mol:0,1mol:0,1mol\)
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
a. \(n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH:
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
\(0,1............0,2........0,2.......0,1.......0,1\) (mol)
a. \(m_{NaCl}=M_{NaCl}.n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=22,4.n_{CO_2}=22,4.0,1=2,24\left(l\right)\)
c. \(m_{HCl}=n_{HCl}.M_{HCl}=0,2.36,5=7,3\left(g\right)\)
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O (1)
NaHCO3 + HCl → NaCl + CO2 + H2O (2)
\(n_{CO_2}=\frac{0,896}{22,4}=0,04\left(mol\right)\)
a) Gọi x,y lần lượt là số mol của Na2CO3 và NaHCO3
Theo PT1,2 ta có: \(\left\{{}\begin{matrix}106x+84y=3,8\\x+y=0,04\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,02\end{matrix}\right.\)
\(n_{Na_2CO_3}=0,02\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,02\times106=2,12\left(g\right)\)
\(n_{NaHCO_3}=0,02\left(mol\right)\Rightarrow m_{NaHCO_3}=0,02\times84=1,68\left(g\right)\)
\(\%m_{Na_2CO_3}=\frac{2,12}{3,8}\times100\%=55,79\%\)
\(\%m_{NaHCO_3}=100\%-55,79\%=44,21\%\)
b) Theo PT1: \(n_{HCl}=2n_{Na_2CO_3}=2\times0,02=0,04\left(mol\right)\)
Theo PT2: \(n_{HCl}=n_{NaHCO_3}=0,02\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,04+0,02=0,06\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,06\times36,5=2,19\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{2,19}{20\%}=10,95\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\frac{10,95}{1,1}=9,95\left(ml\right)\)
c) Theo PT1: \(n_{NaCl}=n_{HCl}=0,04\left(mol\right)\)
Theo pT2: \(n_{NaCl}=n_{HCl}=0,02\left(mol\right)\)
\(\Rightarrow\Sigma n_{NaCl}=0,02+0,04=0,06\left(mol\right)\)
\(m_{NaCl}=0,06\times58,5=3,51\left(g\right)\)
\(C_{M_{NaCl}}=\frac{0,06}{0,0095}=6,32\left(M\right)\)
\(m_{CO_2}=0,04\times44=1,76\left(g\right)\)
Ta có: \(m_{dd}saupư=3,8+10,95-1,76=12,99\left(g\right)\)
\(C\%_{NaCl}=\frac{3,51}{12,99}\times100\%=27,02\%\)
a) CO2 + 2NaOH → Na2CO3 + H2O
b) 3CO2 + 4NaOH → 2NaHCO3 + NaCO3 + H2O
c) P2O5 + 3Na2O → 2Na3PO4
d) 2Na3PO4 + 3BaCl2 → Ba3(PO4)2 + 6NaCl
\(a.Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ n_{NaCl}=n_{HCl}=2.n_{CO_2}=2.\dfrac{448:1000}{22,4}=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,02}=2\left(M\right)\\ b.m_{NaCl}=58,5.0,04=2,34\left(g\right)\\ c.m_{Na_2CO_3}=106.0,02=2,12\left(g\right)\\ \%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\ \%m_{NaCl}=100\%-42,4\%=57,6\%\)
Bài 16 :
\(n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Pt : \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O|\)
1 2 2 1 1
0,02 0,04 0,04 0,02
a) \(n_{HCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
20ml = 0,02l
\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)
b) \(n_{NaCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
⇒ \(m_{NaCl}=0,04.58,5=2,34\left(g\right)\)
c) \(n_{Na2CO3}=\dfrac{0,04.1}{2}=0,02\left(mol\right)\)
⇒ \(m_{Na2CO3}=0,02.106=2,12\left(g\right)\)
\(m_{NaCl}=5-2,12=2,88\left(g\right)\)
0/0Na2CO3 = \(\dfrac{2,12.100}{5}=42,4\)0/0
0/0NaCl = \(\dfrac{2,88.100}{5}=57,6\)0/0
Chúc bạn học tốt
) Gọi n Na2CO3= x ; n Na2So3 =y
=> 106x + 126y= 55 (I)
Ta có: mHCl =36,5 => nHCl= 1
Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O (1)
x 2x 2x x
Na2SO3 + 2HCl --> 2NaCl + SO2 + H2O (2)
y 2y 2y y
=> 2x + 2y = 1 (II)
(I) và (II) => n Na2CO3= 0,4 và n Na2SO3 = 0,1
(1)=> nCO2 = 0,4
(2) => nSO2= 0,1
=> %V CO2 = 80% => %V SO2 = 20%
b) ΣΣ nNaCl= 1=> mNaCl= 58,5
m dd sau pứ= 55 + 250 - 0,4 x 44 - 0,1 x 64= 281(g)
C% dd NaCl = 20,82%