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, \(Na_2O+H_2O\rightarrow2NaOH\)

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:

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): 

\(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,6=0,3\left(mol\right)\\ \rightarrow m_{Na_2SO_4}=0,3.142=42,6\left(g\right)\)

nNa2O=0.1(mol)

pthh:Na2O+H2O->2NaOH

theo pthh:nH2O=nNa2O->nH2O=0.1(mol)->mH2O=0.1*18=1.8(g)

b)Theo pthh:nNaOH=2 nNa2O

->nNaOH=0.1*2=0.2(mol)

CM=0.2:0.2=1(M)

Giải:

a) Số mol Na2O là:

nNa2O = m/M = 15,5/62 = 0,25 (mol)

PTHH: Na2O + H2O -> 2NaOH

---------0,25-----0,25-------0,5---

Nồng độ mol của NaOH là:

C_MNaOH = n/V = 0,5/0,5 = 1 (M)

b)

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

-----------0,5----------0,25------0,25------0,5--

Khối lượng chất tan H2SO4 là:

mctH2SO4 = n.M = 0,25.98 = 24,5 (g)

Khối lượng dd H2SO4 là:

mddH2SO4 = (mct.100)/C% = (24,5.100)/20 = 122,5 (g)

Thể tích dd H2SO4 là:

VH2SO4 = mdd/D = 122,5/1,14 ≃ 107,5 (ml)

c) Nồng độ mol của Na2SO4 là:

C_MNa2SO4 = n/V = 0,25/0,1075 ≃ 2,3 (M)

Vậy ...

a.

\(nNa_2O=\dfrac{m}{M}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

\(P.tr:Na_2O+H_2O\rightarrow2NaOH\)

\(..........1......1.......2\)

\(........0,25.....0,25.....0,5\)

Nồng độ mol của NaOH:

\(C_MNaOH=\dfrac{n}{V}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b)

\(Ptr:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(........2..........1............1.........2\)

\(.......0,5..........0,25.............0,25.......0,5\)

\(mH_2SO_4=n.M=0,25.98=24,5\left(g\right)\)

\(\Rightarrow mddH_2SO_4=\dfrac{mH_2SO_4.100\%}{C\%}=\dfrac{24,5.100\%}{20\%}=122,5\left(g\right)\)

\(\Rightarrow VddH_2SO_4=\dfrac{m}{D}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

Vậy thể tích dung dịch H₂SO₄ cần dùng là 107,46l

c.Đổi:\(107,46ml=0,10746l\)

Nồng độ mol của Na₂SO₄ thu được sau phản ứng là:

\(C_MNa_2SO_4=\dfrac{n}{V}=\dfrac{0,25}{0,10746}\approx2,33\left(M\right)\)

a) PTHH: Na2O + H2O ------> 2NaOH (1)

nNa2O = 0,25 mol

- Theo PTHH(1) : nNaOH = 0,5 mol

=> CM NaOH = 1M

b) PTHH: 2NaOH (0,5) + H2SO4 (0,25) ------> Na2SO4 (0,25) + 2H2O (2)

-Theo PTHH(2): nH2SO4 = 0,25 mol

=> mH2SO4 = 24,5 gam

=> mddH2SO4 = 122,5 gam

=> VddH2SO4 = 122,5/1,14 = 107,456 (ml)

c) - Theo PTHH(2): nNa2SO4 = 0,25 mol

=> CM Na2SO4 = \(\dfrac{0,25}{0,5+0,107456}=0,4116M\)

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

\(Na_2O+H_2O\rightarrow2NaOH\)

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

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

b.

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

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

\(V_{dd_{H_2SO_4}}=\dfrac{0.25}{2}=0.125\left(l\right)\)

\(a.n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ 0,25...................................0,5\left(mol\right)\\ C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ 0,25............0,5..........0,25\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,25}{2}=0,125\left(l\right)\)