\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ a,n_{NaOH}=0,4(mol);n_{Na_2SO_4}=0,2(mol)\\ \Rightarrow \begin{cases} m_{Na_2SO_4}=0,2.142=28,4(g)\\ m_{dd_{NaOH}}=\dfrac{0,4.40}{20\%}=80(g) \end{cases}\\ b,2KOH+H_2SO_4\to K_2SO_4+2H_2O\\ \Rightarrow n_{KOH}=0,4(mol)\\ \Rightarrow m_{dd_{KOH}}=\dfrac{0,4.56}{5,6\%}=400(g)\\ \Rightarrow V_{dd_{KOH}}=\dfrac{400}{1,045}=382,78(ml)\)

Bước 1: nH2SO4 = VH2SO4 . CM H2SO4= 0,2 . 1 = 0,2mol

Bước 2:

PTHH:       2NaOH    +    H2SO4 →  Na2SO4 + H2O

                     2 mol             1 mol     

                    ? mol               0,2mol

m NaOH= n NaOH.MNaOH = 0,4 . (23 + 16 + 1) = 16g

Bước 3: C% = mNaOH : m dd NaOH => mdd NaOH = mNaOH : C% = 16 : 20% = 80g

Câu 16:

PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)

Ta có: \(n_{HCl}=0,25\cdot1,5=0,375\left(mol\right)=n_{KOH}=n_{KCl}\)

\(\Rightarrow\left\{{}\begin{matrix}V_{KOH}=\dfrac{0,375}{2}=0,1875\left(l\right)\\C_{M_{KCl}}=\dfrac{0,375}{0,1875+0,25}\approx0,86\left(M\right)\end{matrix}\right.\)

Câu 18:

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

a) Ta có: \(n_{H_2SO_4}=\dfrac{200\cdot14,7\%}{98}=0,3\left(mol\right)\)

\(\Rightarrow n_{KOH}=0,6\left(mol\right)\) \(\Rightarrow m_{ddKOH}=\dfrac{0,6\cdot56}{5,6\%}=600\left(g\right)\) \(\Rightarrow V_{ddKOH}=\dfrac{600}{10,45}\approx57,42\left(ml\right)\)

b) Theo PTHH: \(n_{K_2SO_4}=0,3\left(mol\right)\) \(\Rightarrow C\%_{K_2SO_4}=\dfrac{0,3\cdot174}{600+200}\cdot100\%=6,525\%\)

a) nH2SO4 = 0,2 . 1 = 0,2 mol

H2SO4  +   2NaOH  -> Na2SO4  + 2H2O

0,2                0,4

mNaOH = 0,4 . 40 = 16g

mddNaOH = \(\frac{16.100\%}{20\%}=80g\)

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

0,4 <---------- 0,2

=> mKOH = 0,4 . 56 = 22,4 g

mddKOH = \(\frac{22,4.100\%}{5,6\%}=400g\)

VddKOH = \(\frac{400}{1,045}=383ml\)

a,b)\(n_{H_2SO_4}=0,2\) theo PT \(2n_{H_2SO_4}=n_{NaOV}=0,4\Rightarrow m_{NaOV}=0,4.4016kg\)

                        \(H_2SO_4+2NaOH\rightarrow H_2Na_2SO_4+H_2\)

\(m_{ddNaOH}=\dfrac{16}{2}=80g\)

c)\(n_{KOH}=2_{n_{H_2SO_4}}=0,4\Rightarrow m_{KOH}=0,3.56-22,4g\)

\(\Rightarrow m_{KOH\left(dd\right)}=\dfrac{22,4}{0,056}=400\Rightarrow V=\dfrac{400}{1,045}=182,8\left(ml\right)\)

a, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)    \(\left(1\right)\)

b, Đổi \(20ml=0,02l\)

\(n_{H_2SO_4}=0,02.1=0,02\left(mol\right)\)

Thep phương trình \(\left(1\right)\) ta được:

\(n_{NaOH}=2n_{H_2SO_4}=2.0,02=0,04\left(mol\right)\\ \Rightarrow m_{NaOH}=0,04.40=1,6\left(g\right)\\ \Rightarrow m_{dd NaOH}=\dfrac{1,6}{20}.100=8\left(g\right)\)

c, \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)    \(\left(2\right)\)

Theo phương trình \(\left(2\right)\):

\(n_{KOH}=2n_{H_2SO_4}=2.0,02=0,04\left(mol\right)\\ \Rightarrow m_{KOH}=0,04.56=2,24\left(g\right)\\ \Rightarrow m_{dd KOH}=\dfrac{2,24}{5,6}.100=40\left(g\right)\\ \Rightarrow V_{dd KOH}=\dfrac{40}{1.045}=38,3\left(ml\right)\)

\(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

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

Theo PT: \(n_{KOH}=2n_{H_2SO_4}=0,3\left(mol\right)\)

\(\Rightarrow m_{KOH}=0,3.56=16,8\left(g\right)\)

\(\Rightarrow m_{ddKOH}=\dfrac{16,8}{5,6\%}=300\left(g\right)\)

\(\Rightarrow V_{ddKOH}=\dfrac{300}{10,45}\approx28,71\left(ml\right)\)

Tính thể tích K₂SO₄ giúp em luôn đc ko ạ

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

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

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

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

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

Dd HCl có CM không bạn?

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

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

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

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