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Câu 1:
a, \(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
Câu 3: \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
\(n_{CaCO_3}=n_{CO_2}=0,4\left(mol\right)\Rightarrow m_{CaCO_3}=0,4.100=40\left(g\right)\)
Câu 4: \(n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)
PT: \(CuSO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+CuCl_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\), ta được CuSO4 dư.
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)
a, \(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=0,25.2=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,5\left(ml\right)\)
a, \(n_{Na_2O}=\dfrac{7,75}{62}=0,125\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,125 0,25
b, \(C_{M_{ddNaOH}}=\dfrac{0,25}{0,25}=1M\)
c,
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,25 0,125
\(m_{ddH_2SO_4}=\dfrac{0,125.98.100}{20}=61,25\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{61,25}{1,14}=53,728\left(ml\right)\)
1. \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,25 0,5
\(C_{M_{ddNaOH}}=\dfrac{0,5}{0,5}=1M\)
2.
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,5 0,25
\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\)
\(V_{ddH_2SO_4}=\dfrac{122,5}{1,14}=107,456\left(ml\right)\)
\(n_{Na2O}=\dfrac{m_{Na2O}}{M_{Na2O}}=0,25\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,25 mol - 0,25 mol - 0,5 mol
a) \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{V_{NaOH}}=1\left(M\right)\)
b) \(H_2SO_4+2NaOH\rightarrow Na_2SO4+2H_2O\)
0,25 mol - 0,5 mol - 0,25 mol - 0,5 mol
\(m_{ctH2SO4}=n_{H2SO4}.M_{H2SO4}=24,5\left(g\right)\)
\(C_{\%_{H2SO4}}=\dfrac{m_{ctH2SO4}}{m_{ddH2SO4}}.100\%\)
\(\Rightarrow m_{ddH2SO4}=\dfrac{m_{ctH2SO4}.100\%}{C_{\%_{H2SO4}}}=122,5\left(g\right)\)
\(D_{H2SO4}=\dfrac{m_{ddH2SO4}}{V_{H2SO4}}\Rightarrow V_{H2SO4}=\dfrac{m_{ddH2SO4}}{D_{H2SO4}}\approx107,46\left(ml\right)\)
a)Gọi C% của dd sau khi trộn là x (%)
Ta có sơ đồ đường chéo:
50g NaOH 8% ........................... 20-x
................ x (%)............
450g NaOH 20%........................... x-8
=> \(\dfrac{50}{250}=\dfrac{20-x}{x-8}\)<=> x = 18,8 %
b) CM = \(\dfrac{18,8.1,1}{40}\)= 0,517 (mol/lít)
mNaOH = \(\dfrac{\left(45+450\right).18,8}{100}\)100 = 94 (g) => nNaOH = \(\dfrac{94}{40}\) = 2,35(mol)
=> Vdd = \(\dfrac{2,35}{0,517}\)≈ 4,55 (l)
a) PTHH: Na2O + H20 -> 2NaOH
số mol Na20 = 0,25 (mol)
=> số mol NaOH = 0,5 mol.
Nôngd độ mol NaOH = 0,5 / 0,5 = 1 M
b) PTHH: H2SO4 + 2NaOH -> Na2SO4 + 2H2O
số mol H2SO4 = 1/2 số mol NaOH = 0,25 mol
C% H2SO4 = mH2SO4 / m ddH2SO4 . 100%
=> m ddH2SO4= 122,5 g
D=m/V => V= 107,5 ml