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\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ a,n_{NaOH}=0,4(mol);n_{Na_2SO_4}=0,2(mol)\\ \Rightarrow \begin{cases} m_{Na_2SO_4}=0,2.142=28,4(g)\\ m_{dd_{NaOH}}=\dfrac{0,4.40}{20\%}=80(g) \end{cases}\\ b,2KOH+H_2SO_4\to K_2SO_4+2H_2O\\ \Rightarrow n_{KOH}=0,4(mol)\\ \Rightarrow m_{dd_{KOH}}=\dfrac{0,4.56}{5,6\%}=400(g)\\ \Rightarrow V_{dd_{KOH}}=\dfrac{400}{1,045}=382,78(ml)\)
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
nNaOH=0,2.21=0,4mol.nNaOH=0,2.21=0,4mol.
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: \(n_{KOH}=2n_{H_2SO_4}=0,3\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,3.56=16,8\left(g\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{16,8}{5,6\%}=300\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{300}{10,45}\approx28,71\left(ml\right)\)
PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO+2H_2O\)
Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{KOH}=0,4\left(mol\right)\\n_{K_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddKOH}=\dfrac{\dfrac{0,4\cdot56}{6\%}}{1,048}\approx356,2\left(ml\right)\\C_{M_{K_2SO_4}}=\dfrac{0,2}{0,2+0,3562}\approx0,36\left(M\right)\end{matrix}\right.\)
\(n_{H_2SO_4}=1.0,2=0,2\left(mol\right)\\ H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\\ 0,2.........0,4........0,2.......0,2\left(mol\right)\\ a.m_{ddKOH}=\dfrac{0,4.56.100}{6}=\dfrac{1120}{3}\left(g\right)\\ V_{ddKOH}=\dfrac{\dfrac{1120}{3}}{1,048}=\dfrac{140000}{393}\left(ml\right)\approx0,356\left(l\right)\)
\(b.C_{MddK_2SO_4}=\dfrac{0,2}{\dfrac{140000}{393}+0,2}\approx0,00056\left(M\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{200}.100\%=14,7\%\)
=> \(m_{H_2SO_4}=29,4\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
a. PTHH: 2KOH + H2SO4 ---> K2SO4 + 2H2O
Theo PT: \(n_{KOH}=2.n_{H_2SO_4}=2.0,3=0,6\left(mol\right)\)
=> \(m_{KOH}=0,6.56=33,6\left(g\right)\)
Ta có: \(C_{\%_{KOH}}=\dfrac{33,6}{m_{dd_{KOH}}}.100\%=5,6\%\)
=> \(m_{dd_{KOH}}=600\left(g\right)\)
Theo đề, ta có:
\(D=\dfrac{600}{V_{dd_{KOH}}}=10,45\)(g/ml)
=> \(V_{dd_{KOH}}=57,42\left(ml\right)\)
b. Ta có: \(m_{dd_{K_2SO_4}}=200+33,6=233,6\left(g\right)\)
Theo PT: \(n_{K_2SO_4}=n_{H_2SO_4}=0,3\left(mol\right)\)
=> \(m_{K_2SO_4}=0,3.174=52,2\left(g\right)\)
=> \(C_{\%_{K_2SO_4}}=\dfrac{52,2}{233,6}.100\%=22,35\%\)
Đề cho 200 ml dd axit nồng độ bao nhiêu bạn nhỉ?
nó ko có bn ơi chỉ để nhiêu đó thôi nha