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Na2O+H2O->2NaOH
0,3----------------0,6
2NaOH+H2SO4->Na2SO4+H2O
0,6----------------------0,3 mol
n Na2O=\(\dfrac{18,6}{62}\)=0,3 mol
=>m NaOH= 0,6.40=24g
=> m Na2SO4=0,3.142=42,6g
\(n_{Na_2O}=\dfrac{15.5}{62}=0.25\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.25.......................0.5\)
\(C_{M_{NaOH}}=\dfrac{0.5}{0.5}=1\left(M\right)\)
b.
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(0.5..............0.25\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.25}{2}=0.125\left(l\right)\)
\(a.n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ 0,25...................................0,5\left(mol\right)\\ C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ 0,25............0,5..........0,25\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,25}{2}=0,125\left(l\right)\)
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, \(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
\(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
a, \(n_{Ca}=\dfrac{12}{40}=0,3\left(mol\right)\)
PT: \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
Theo PT: \(n_{H_2}=n_{Ca}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Ca\left(OH\right)_2}=n_{Ca}=0,3\left(mol\right)\Rightarrow m_{Ca\left(OH\right)_2}=0,2.74=22,2\left(g\right)\)
c, \(n_{Fe_3O_4}=\dfrac{8,4}{232}=\dfrac{21}{580}\left(mol\right)\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Xét tỉ lệ: \(\dfrac{\dfrac{21}{580}}{1}< \dfrac{0,3}{4}\), ta được H2 dư.
Theo PT: \(n_{Fe}=3n_{Fe_3O_4}=\dfrac{63}{580}\left(mol\right)\Rightarrow m_{cr}=m_{Fe}=\dfrac{63}{580}.56=\dfrac{882}{145}\left(g\right)\)
a, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 4Na + O2 ---to→ 2Na2O
Mol: 0,2 0,1
PTHH: Na2O + 2HCl → 2NaCl + H2O
Mol: 0,1 0,2
b, \(m_{Na_2O}=0,1.62=6,2\left(g\right)\)
c, \(V=V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)
a, PTHH:
Na2O + H2O ---> 2NaOH (1)
2NaOH + H2SO4 ---> Na2SO4 + 2H2O (2)
b, \(n_{Na_2O}=\dfrac{18,6}{62}=0,3\left(mol\right)\)
Theo pthh (1): \(n_{NaOH}=2n_{Na_2O}=2.0,3=0,6\left(mol\right)\)
=> \(m_{NaOH}=0,6.40=24\left(g\right)\)
c, \(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
LTL (2): \(\dfrac{0,6}{2}< 0,5\rightarrow\) H2SO4 dư
Theo pthh (2):
\(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,6=0,3\left(mol\right)\\ \rightarrow m_{Na_2SO_4}=0,3.142=42,6\left(g\right)\)