nH2SO4 = 0.2*1=0.2 mol

2NaOH + H2SO4 --> Na2SO4 + H2O

0.4________0.2

mNaOH = 0.4*40=16g

2KOH + H2SO4 --> K2SO4 + H2O

0.4______0.2

mKOH= 0.4*56=22.4g

mddKOH = 22.4*100/5.6=400g

VddKOH = 400/1.045=382.77ml

\(n_{H_2SO_4}=0,2\times1=0,2\left(mol\right)\)

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (1)

a) Theo PT1: \(n_{NaOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,4\times40=16\left(g\right)\)

b) H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O (2)

Theo PT2: \(n_{KOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{KOH}=0,4\times56=22,4\left(g\right)\)

\(\Rightarrow m_{ddKOH}=\frac{22,4}{5,6\%}=400\left(g\right)\)

\(\Rightarrow V_{ddKOH}=\frac{400}{1,045}=382,78\left(ml\right)\)

a) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

b)

n H2SO4 = 0,03.1 = 0,03(mol)

n NaOH = 2n H2SO4 = 0,06(mol)

=> CM NaOH = 0,06/0,05 = 1,2M

c) $H_2SO_4 + 2KOH \to K_2SO_4 + 2H_2O$

n KOH = 2n H2SO4 = 0,06(mol)

=> m KOH = 0,06.56 = 3,36 gam

=> m dd KOH = 3,36/5,6% = 60(gam)

=> V dd KOH = m/D = 60/1,045 = 57,42(ml)

a) H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

b) \(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

PTHH: H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

                 0,2---->0,4

=> m_NaOH = 0,4.40 = 16 (g)

=> \(m_{dd.NaOH}=\dfrac{16.100}{20}=80\left(g\right)\)

c)

PTHH: H₂SO₄ + 2KOH --> K₂SO₄ + 2H₂O

                0,2---->0,4

=> m_KOH = 0,4.56 = 22,4 (g)

=> \(m_{dd.KOH}=\dfrac{22,4.100}{5,6}=400\left(g\right)\)

=> \(V_{dd.KOH}=\dfrac{400}{1,045}=382,775\left(ml\right)\)

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ pthh:H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) 
          0,2                   0,4 
\(m_{\text{ }NaOH}=0,4.40=16g\\ m_{\text{dd}NaOH}=\dfrac{16.100}{20}=80g\) 
\(pthh:H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\) 
            0,2             0,4 
\(m_{KOH}=0,4.56=22,4g\\ m_{\text{dd}KOH}=\dfrac{22,4.100}{5,6}=400g\\ V_{\text{dd}}=\dfrac{400}{1,045}=382,7ml\)
\(V_{dd}=\dfrac{22,4}{1,045}=21,4354ml\)

\(n_{H2SO4}=0,02.1=0,02\left(mol\right)\)

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

pư..............0,02..........0,04..............0,02...........0,04 (mol)

\(\Rightarrow m_{NaOH}=0,04.40=1,6\left(g\right)\)

\(\Rightarrow m_{ddNaOH\left(20\%\right)}=\dfrac{1,6}{20\%}=8\left(g\right)\)

PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)

pư............0,02............0,04............0,02..........0,04 (mol)

\(\Rightarrow m_{KOH}=0,04.56=2,24\left(g\right)\)

\(\Rightarrow m_{ddKOH\left(5,6\%\right)}=\dfrac{2,24}{5,6\%}=40\left(g\right)\)

\(\Rightarrow V_{KOH}=\dfrac{40}{1,045}\approx38,28\left(ml\right)\)

Vậy..............

nH2SO4=0,1(mol) 

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

b) 0,2___________0,1________0,1(mol)

mNaOH=0,2.40=8(g)

=>mddNaOH=(8.100)/25= 32(g)

c) mNa2SO4=0,1.142=14,2(g)

d) PTHH: 2 KOH + H2SO4 -> K2SO4 +2 H2O

nKOH=0,2(mol) => mKOH=0,2.56=11,2(g)

=> mddKOH=(11,2.100)/8=140(g)

=> VddKOH= 140/1,085=129,03(ml)

Chúc em học tốt!

nH2SO4=0,02.1=0,02(ol)

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

0,04____________0,02____0,02(mol)

mNaOH=0,04.40= 1,6(g)

=>mddNaOH= (1,6.100)/20= 8(g)

b) PTHH: H2SO4 + 2 KOH -> K2SO4 + 2 H2O

0,2____________0,04(mol)

=>mKOH=0,04.56=2,24(g)

=>mddKOH= (2,24.100)/5,6=40(g)

=>VddKOH= mddKOH/DddKOH= 40/1,045=38,278(ml)

PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO+2H_2O\)

Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KOH}=0,4\left(mol\right)\\n_{K_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddKOH}=\dfrac{\dfrac{0,4\cdot56}{6\%}}{1,048}\approx356,2\left(ml\right)\\C_{M_{K_2SO_4}}=\dfrac{0,2}{0,2+0,3562}\approx0,36\left(M\right)\end{matrix}\right.\)   

\(n_{H_2SO_4}=1.0,2=0,2\left(mol\right)\\ H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\\ 0,2.........0,4........0,2.......0,2\left(mol\right)\\ a.m_{ddKOH}=\dfrac{0,4.56.100}{6}=\dfrac{1120}{3}\left(g\right)\\ V_{ddKOH}=\dfrac{\dfrac{1120}{3}}{1,048}=\dfrac{140000}{393}\left(ml\right)\approx0,356\left(l\right)\)

\(b.C_{MddK_2SO_4}=\dfrac{0,2}{\dfrac{140000}{393}+0,2}\approx0,00056\left(M\right)\)

Đổi 100ml = 0,1 lít

Ta có: \(C_{M_{KOH}}=\dfrac{n_{KOH}}{0,1}=0,5M\)

=> \(n_{KOH}=0,05\left(mol\right)\)

a. PTHH: 2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O

b. Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{KOH}=\dfrac{1}{2}.0,05=0,025\left(mol\right)\)

Vì H₂SO₄ là chất lỏng nên thể tích bằng số mol của chính nó.

c. PTHH: KOH + HCl ---> KCl + H₂O

Theo PT: \(n_{HCl}=n_{KOH}=0,05\left(mol\right)\)

=> \(m_{HCl}=0,05.36,5=1,825\left(g\right)\)

Ta có; \(C_{\%_{HCl}}=\dfrac{1,825}{m_{dd_{HCl}}}.100\%=20\%\)

=> \(m_{dd_{HCl}}=9,125\left(g\right)\)