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\(Bài.4:\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\K_2O+H_2O\rightarrow2KOH\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Na}=n_{NaOH}=2.0,1=0,2\left(mol\right)\\ m_{Na}=0,2.23=4,6\left(g\right)\\ \Rightarrow m_{K_2O}=9,3-4,6=4,7\left(g\right)\Rightarrow n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\\ n_{KOH}=0,05.2=0,1\left(mol\right)\\ m_{ddA}=m_X+m_{H_2O}-m_{H_2}=9,3+70,9-0,1.2=80\left(g\right)\\ C\%_{ddNaOH}=\dfrac{0,2.40}{80}.100=10\%\\ C\%_{ddKOH}=\dfrac{0,1.56}{80}.100=7\%\)
\(Bài.5\\R_2O+H_2O\rightarrow2ROH\\m_{ddROH}=23,5+176,5=200\left(g\right)\\ m_{ROH}=200.14\%=28\left(g\right)\\ Ta.có:28.\left(2M_R+16\right)=23,5.\left(2M_R+34\right)\\ \Leftrightarrow 9M_R=351\\ \Leftrightarrow M_R=39\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(I\right):Kali\left(K=39\right)\\ \Rightarrow CTHH.oxit:K_2O\)
1)
$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{dd} = 6,2 + 193,8 = 200(gam) \Rightarrow C\%_{NaOH} = \dfrac{0,2.40}{200}.100\% = 4\%$
2)
$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,5(mol) \Rightarrow C_{M_{KOH}} = \dfrac{0,5}{0,5} = 1M$
3) $n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} =0,8M$
4)
$Na_2SO_3 + 2HCl \to 2NaCl +S O_2 + H_2O$
Theo PTHH :
$n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$
5) $n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
Theo PTHH :
$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
\(K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=\dfrac{150.11,2\%}{56}=0,3\left(mol\right)\\ n_{K_2O}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ m_{K_2O}=0,15.94=14,1\left(g\right)\\ \Rightarrow m=14,1\left(g\right)\)
PTHH: \(K_2O+H_2O\rightarrow2KOH\)
a) \(n_{K_2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\)
\(n_{KOH}=2n_{K_2O}=0,5\left(mol\right)\)
\(m_{KOH}=0,5.56=28\left(g\right)\)
b) \(C_{M_{ddKOH}}=\dfrac{n}{V}=\dfrac{0,5}{0,4}=1,25M\)
Câu 1:
\(n_{K2O}=\frac{9,4}{39.2+16}=0,1\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
0,1_____________0,2
\(C\%_{KOH}=\frac{0,2.\left(39+17\right)}{150,6+9,4}.100\%=7\%\)
\(KOH+HCl\rightarrow KCl+H_2O\)
0,2______0,2__________________
\(\Rightarrow V_{dd_{HCl}}=\frac{0,2}{0,5}=0,5\left(l\right)\)
Câu 2:
a, \(n_{K2O}=\frac{23,5}{39.2+16}=0,25\left(mol\right)\)
\(2n_{K2O}=n_{KOH}\Rightarrow n_{KOH}=0,25.2=0,5\left(mol\right)\)
\(C\%_{KOH}=\frac{0,5.\left(39+17\right)}{176,5+23,5}.100\%=14\%\)
b, \(n_{KOH}=2n_{K2SO4}\Rightarrow n_{K2SO4}=\frac{0,5}{2}=0,25\)
\(n_{H2SO4}=n_{K2SO4}=0,25\)
\(m_{dd_{H2SO4}}=\frac{0,25.98}{20\%}=122,5\left(g\right)\)
c,
mdd sau phản ứng=mddA+mddH2SO4
m dd sau phản ứng \(=23,5+176,5+122,5=322,5\)
\(C\%_{K2SO4}=\frac{0,25.\left(39.2+32+16.4\right)}{322,5}.100\%=13,49\%\)
\(m_{ddNaCl}=25+100=125\left(g\right)\\ C\%_{ddNaCl}=\dfrac{25}{125}.100=20\%\\ \Rightarrow ChọnB\)
a) PTHH: \(K_2O+H_2O\rightarrow2KOH\)
Ta có: \(n_{KOH}=2n_{K_2O}=2\cdot\dfrac{35,25}{94}=0,75\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,75}{0,75}=1\left(M\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{KOH}=0,75\left(mol\right)\\n_{CO_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo muối trung hòa
PTHH: \(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)
Theo PTHH: \(n_{K_2CO_3}=0,375\left(mol\right)\) \(\Rightarrow m_{K_2CO_3}=0,375\cdot138=51,75\left(g\right)\)
c) PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
Theo PTHH: \(n_{H_2SO_4}=0,375\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,375\cdot98}{60\%}=61,25\left(g\right)\) \(\Rightarrow V_{ddH_2SO_4}=\dfrac{61,25}{1,5}\approx40,83\left(ml\right)\)
\(n_{K_2O}=\dfrac{23,5}{94}=0,25mol\\ K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=2.0,25=0,5mol\\ C_{\%KOH}=\dfrac{0,5.56}{23,5+176,5}\cdot100\%=14\%\)