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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.250ml=0,25l ; nHCl=0,25.1,5=0,375mol
KOH+HCl->KCl+H2O
1mol 1mol 1mol
0,375 0,375 0,375
VKOh=0,375/2=0,1875l
b.CM KCL=0,375/0,25=1,5M
c.NaOH+HCL=NaCl+H2O
1mol 1mol
0,375 0,375
mdd NaOH=0,375.40.100/10=150g
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PT: \(n_{KOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)
\(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)
PT: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
Theo PT: \(n_{KOH}=2n_{H_2SO_4}=0,3\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,3.56=16,8\left(g\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{16,8}{5,6\%}=300\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{300}{10,45}\approx28,71\left(ml\right)\)
a)
\(n_{Na_2O}=\dfrac{31}{62}=0,5\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
______0,5--------------->1
=> \(C_{M\left(NaOH\right)}=\dfrac{1}{0,5}=2M\)
b)
PTHH: H2SO4 + 2NaOH --> Na2SO4 + 2H2O
______0,5<---------1
=> mH2SO4 = 0,5.98 = 49(g)
=> \(m_{dd\left(H_2SO_4\right)}=\dfrac{49.100}{20}=245\left(g\right)\)
=> \(V_{dd\left(H_2SO_4\right)}=\dfrac{245}{1,14}=214,912\left(ml\right)\)
\(n_{Na_2O}=\dfrac{31}{62}=0,5(mol)\\ a,Na_2O+H_2O\to 2NaOH\\ \Rightarrow n_{NaOH}=1(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1}{0,5}=2M\\ b,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,5(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,5.98}{20\%}=245(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{245}{1,14}=214,91(ml)\)
PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO+2H_2O\)
Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{KOH}=0,4\left(mol\right)\\n_{K_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddKOH}=\dfrac{\dfrac{0,4\cdot56}{6\%}}{1,048}\approx356,2\left(ml\right)\\C_{M_{K_2SO_4}}=\dfrac{0,2}{0,2+0,3562}\approx0,36\left(M\right)\end{matrix}\right.\)
\(n_{H_2SO_4}=1.0,2=0,2\left(mol\right)\\ H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\\ 0,2.........0,4........0,2.......0,2\left(mol\right)\\ a.m_{ddKOH}=\dfrac{0,4.56.100}{6}=\dfrac{1120}{3}\left(g\right)\\ V_{ddKOH}=\dfrac{\dfrac{1120}{3}}{1,048}=\dfrac{140000}{393}\left(ml\right)\approx0,356\left(l\right)\)
\(b.C_{MddK_2SO_4}=\dfrac{0,2}{\dfrac{140000}{393}+0,2}\approx0,00056\left(M\right)\)