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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: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(NaOH+CO_2\rightarrow NaHCO_3\)
a) \(n_{CO_2}=\dfrac{7}{22,4}=\dfrac{5}{16}\left(mol\right)\)
\(n_{NaOH}=\dfrac{5,8}{40}=0,145\left(mol\right)\)
PTHH: 2NaOH + CO2 --> Na2CO3 + H2O
______0,145->0,0725-->0,0725
\(Na_2CO_3+CO_2+H_2O->2NaHCO_3\)
_0,0725->0,0725------------->0,145
=> Muối thu được là NaHCO3: 0,145 mol
b) CO2 dư
\(n_{CO_2\left(dư\right)}=\dfrac{5}{16}-0,145=0,1675\left(mol\right)\)
a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: CO2 + 2NaOH → Na2CO3 + H2O
Mol: 0,15 0,3 0,15
\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,2}=1,5M\)
b) Na2CO3: natri cacbonat
\(m_{Na_2CO_3}=0,15.106=15,9\left(g\right)\)
c)
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,15 0,075
\(V_{ddH_2SO_4}=\dfrac{0,075}{1}=0,075\left(l\right)=75\left(ml\right)\)
a)CO2+2NaOH----->Na2CO3+H2O
b)n\(_{CO2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
Theo pthh
n\(_{NaOH}=2n_{CO2}=0,1\left(mol\right)\)
V\(_{NaOH}=\frac{0,1}{1}=0,1\left(l\right)=100ml\)
c) Theo pthh
n\(_{Na2CO3}=n_{CO2}=0,05\left(mol\right)\)
C\(_{M\left(Na2CO3\right)}=\frac{0,05}{0,1}=0,5\left(M\right)\)
a/ \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
PTHH: 3Mg +8HNO3 → 3Mg(NO3)2 + 2NO + 4H2O
Mol: 0,5 4/3 1/3
b, \(V_{NO}=\dfrac{1}{3}.22,4=\dfrac{112}{15}\approx7,46\left(l\right)\)
c, \(V_{ddHNO_3}=\dfrac{\dfrac{4}{3}}{0,5}=\dfrac{8}{3}\approx2,667\left(l\right)\)
Bài 1:
a) PTHH: 2 NaOH + SO2 -> Na2SO3 + H2O
nSO2= 5,6/22,4= 0,25(mol)
b) nNa2SO3= nSO2 = 0,25(mol)
=> mNa2SO3= 0,25. 126= 31,5(g)
c) nNaOH= 2.0,25= 0,5(mol)
=> VddNaOH = 0,5/1,25= 0,4 (l)
d) VddNa2SO3= VddNaOH= 0,4(l)
=> \(C_{MddNa_2SO_3}=\dfrac{0,25}{0,4}=0,625\left(M\right)\)
Bài 2:
a) PTHH: CO2 + Ca(OH)2 -> CaCO3 (trắng) + H2O
mCa(OH)2= (7,4.100)/100= 29,6(g)
=> nCa(OH)2 = 29,6/ 74= 0,4(mol)
b) => nCaCO3= nCO2= nCa(OH)2= 0,4(mol)
=> m(muối)= mCaCO3= 100.0,4= 40(g)
c) V(CO2, đktc)= 0,4.22,4= 8,96(l)
\(a)2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\\ b)n_{CO_2}=\dfrac{1,1}{44}=0,025mol\\ n_{NaOH}=0,025.2=0,05mol\\ V_{ddNaOH}=\dfrac{0,05}{1,2}=\dfrac{1}{24}\approx0,04l\)