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a, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
b, \(n_{H_2SO_4}=0,2.1,5=0,3\left(mol\right)\)
Theo PT: nNa2SO4 = nH2SO4 = 0,3 (mol) ⇒ mNa2SO4 = 0,3.142 = 42,6 (g)
nNaOH = 2nH2SO4 = 0,6 (mol) ⇒ mNaOH = 0,6.40 = 24 (g)
c, \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,6}{0,4}=1,5\) → Pư tạo NaHCO3 và Na2CO3.
PT: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(NaOH+CO_2\rightarrow NaHCO_3\)
1.
a, \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: CO2 + 2NaOH → Na2CO3 + H2O
Mol: 0,05 0,1
b, \(C_{M_{ddNaOH}}=\dfrac{0,1}{0,1}=1M\)
2.
a, \(m_{HCl}=200.7,3\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: 0,2 0,4 0,2
b,\(m_{CuO}=0,2.80=16\left(g\right)\)
c, \(C\%_{ddCuCl_2}=\dfrac{0,2.135.100\%}{16+200}=12,5\%\)
Bài 7:
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{NaOH}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối
PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
a_______2a__________a (mol)
\(CO_2+NaOH\rightarrow NaHCO_3\)
b_______b__________b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}a+b=0,15\\2a+b=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{CO_2}+m_{ddNaOH}=0,15\cdot44+200\cdot1,25=256,6\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2CO_3}=\dfrac{0,05\cdot106}{256,6}\cdot100\%\approx2,1\%\\C\%_{NaHCO_3}=\dfrac{0,1\cdot72}{256,6}\cdot100\%\approx2,8\%\end{matrix}\right.\)
Bài 8:
PTHH: \(RCO_3+2HNO_3\rightarrow R\left(NO_3\right)_2+CO_2\uparrow+H_2O\)
Giả sử \(n_{RCO_3}=1\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{HNO_3}=2\left(mol\right)\\n_{R\left(NO_3\right)_2}=1\left(mol\right)=n_{CO_2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ddHNO_3}=\dfrac{2\cdot63}{20\%}=630\left(g\right)\\m_{R\left(NO_3\right)_2}=R+124\left(g\right)\\m_{CO_2}=44\left(g\right)\end{matrix}\right.\) \(\Rightarrow C\%_{R\left(NO_3\right)_2}=\dfrac{124+R}{R+60+630-44}=0,26582\)
\(\Leftrightarrow R=65\) (Kẽm) \(\Rightarrow\) CTHH của muối cacbonat là ZnCO3
n CO2=\(\dfrac{6,72}{22,4}\)=0,3 mol
n NaOH=\(2.0,225\)=0,45 mol
T=\(\dfrac{0,3}{0,45}\)=\(\dfrac{2}{3}\)
=>Tạo ra 2 muối NaHCO3 và Na2CO3
2NaOH+CO2->Na2CO3+H2O
0,45-------0,225-------0,225
Na2CO3+H2O+CO2->2NaHCO3
0,075------------0,075--------0,15 mol
=>m NaHCO3=0,15.84=12,6g
=>m Na2CO3= 0,15.106=15,9g
Bài 3 :
a, \(NaOH+HCl\rightarrow NaCl+H_2O\)
b, \(m_{NaOH}=\frac{40.20}{100}=8\left(g\right)\)
\(\rightarrow n_{NaOH}=0,2\left(mol\right)\)
Theo pt: nHCl= nNaOH= 0,2 mol
\(\rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(\rightarrow m_{dd}=29,2\left(g\right)\)
c, \(m_{NaCl}=0,2.\left(23+35,5\right)=11,7\left(g\right)\)
\(\rightarrow m_{dd}=29,2+40=69,2\left(g\right)\)
\(\rightarrow\%_{NaCl}=\frac{11,7}{69,2}=16,9\%\)
\(a/ CO_2+2NaOH \to Na_2CO_3+H_2O\\ n_{CO_2}=0,1(mol)\\ b/\\ n_{NaOH}=0,1.2=0,2(mol)\\ CM_{NaOH}=\frac{0,2}{0,2}=1M\\ c/\\ n_{Na_2CO_3}=n_{CO_2}=0,1(mol)\\ m_{Na_2CO_3}=0,1.106==10,6(g)$\)
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