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

a. \(n_{Na_2O}=\dfrac{18,6}{62}=0,3\left(mol\right)\)

PTHH : Na₂O + H₂O -> 2NaOH

            0,3                      0,6

b. \(m_{NaOH}=0,6.40=24\left(g\right)\)

c. PTHH : 2NaOH + H₂SO₄ -> Na₂SO₄ + 2H₂O

                   0,6                            0,3

\(m_{Na_2SO_4}=0,3.142=42,6\left(g\right)\)

\(n_{Na_2O}=\dfrac{15.5}{62}=0.25\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.25.......................0.5\)

\(C_{M_{NaOH}}=\dfrac{0.5}{0.5}=1\left(M\right)\)

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, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

Giải hộ mình vs ạaaa

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 ---t^o→ 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, \(V=V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)