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bổ sung:
\(m_{dd}=400+0,2.56=411,2\left(g\right)\)
\(C\%_{H2SO4}dư=\dfrac{\left(0,2-0,1\right)98}{411,2}.100\%=2,38\%\)
\(n_{K2SO4}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)
\(C\%_{K2SO4}=\dfrac{0,1.174}{411,2}.100\%=4,23\%\)
Chúc bạn học tốt
\(m_{H_2SO_4}=\dfrac{19,6.100}{100}=19,6\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH: 2KOH + H2SO4 ---> K2SO4 + 2H2O
0,4<-----0,2--------->0,2
\(\rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\\ m_{dd\left(sau.pư\right)}=400+100=500\left(g\right)\\ m_{K_2SO_4}=174.0,2=34,8\left(g\right)\\ \rightarrow C\%_{K_2SO_4}=\dfrac{34,8}{500}.100\%=6,96\%\)
\(n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2mol\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
0,4 0,2 0,2 ( mol )
\(m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400g\)
\(C\%_{K_2SO_4}=\dfrac{0,2.174}{100+400}.100=6,96\%\)
VD1:\(m_{dd}=m_{NaOH}+m_{H_2O}=30+200=230\left(g\right)\)
\(C\%=\dfrac{m_{NaOH}}{m_{dd}}.100\%=\dfrac{30}{230}.100\%\approx13,04\%\)
VD2: \(C\%=\dfrac{m_{NaOH}}{m_{dd}}.100\%=\dfrac{60}{300}.100\%=20\%\)
nK2O = 9.4/94=0.1 mol
K2O + H2O --> 2KOH
0.1____________0.2
mKOH = 0.2*56=11.2g
Dung dịch sau phản ứng làm quỳ tím hóa xanh
C%KOH = 11.2/200*100% = 5.6%
\(n_{CO_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(n_{KOH}=\dfrac{400\cdot5.6\%}{56}=0.4\left(mol\right)\)
\(n_{K_2CO_3}=a\left(mol\right),n_{KHCO_3}=b\left(mol\right)\)
\(2KOH+CO_2\rightarrow K_2CO_3+H_2O\)
\(KOH+CO_2\rightarrow KHCO_3\)
\(\left\{{}\begin{matrix}2a+b=0.4\\a+b=0.3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.2\end{matrix}\right.\)
\(K_2CO_3+BaCl_2\rightarrow BaCO_3+2KCl\)
\(0.1...............................0.1\)
\(m_{BaCO_3}=0.1\cdot197=19.7\left(g\right)\)
Ta có : \(n_{CO2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{KOH}=\dfrac{400.5,6}{56.100}=0,4\left(mol\right)\)
\(\dfrac{n_{KOH}}{n_{CO2}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\Rightarrow\)Tạo 2 muối
\(CO2+2KOH\rightarrow K2CO3+H2O\)
x---------->2x-------->x(mol)
\(CO2+KOH\rightarrow KHCO3\)
y-------->y------------->(mol)
Theo bài ta có HPT\(\left\{{}\begin{matrix}x+y=0,3\\2x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(K2CO3+BaCl2\rightarrow BaCO3+2KCl\)
0,1--------------------------->0,1(mol)
\(\Rightarrow m=m_{BaCO3}=0,1.197=19,7\left(g\right)\)
Chúc bạn học tốt ^.^
\(n_{K_2O}=\dfrac{2,35}{94}=0,025\left(mol\right)\\ K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=0,025.2=0,05\left(mol\right)\\ V_{ddKOH}=V_{ddH_2O}=400\left(ml\right)=0,4\left(l\right)\\ C_{MddKOH}=\dfrac{0,05}{0,4}=0,125\left(M\right)\)
400ml = 0,4l
\(n_{K2O}=\dfrac{2,35}{94}=0,025\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,025 0,05
\(n_{KOH}=\dfrac{0,025.2}{1}=0,05\left(mol\right)\)
\(C_{M_{ddKOH}}=\dfrac{0,05}{0,4}=0,125\left(M\right)\)
Chúc bạn học tốt
Khối lượng KOH cần hòa tan là:
\(m_{KOH}=\frac{200.20}{100}=40\left(g\right)\)
Khối lượng H2O cần thêm: \(200-40=160\left(g\right)\)
Vậy cần hòa tan 40 g K2O vào 160g H2O để thu được 200g dd KOH 20 %
a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)
b, \(n_{KOH\left(trong300mlA\right)}=0,3.0,3=0,09\left(mol\right)\)
Gọi: VH2O = a (l)
\(\Rightarrow C_{M_A}=0,2=\dfrac{0,09}{a+0,3}\Rightarrow a=0,15\left(l\right)=150\left(ml\right)\)
nK2O = 9.4/94 = 0.1 (mol)
mKOH = 200*5.6/100 = 11.2 (g)
nKOH = 11.2/56 = 0.2 (mol)
K2O + H2O => 2KOH
0.1.........................0.2
nKOH = 0.2 + 0.2 = 0.4 (mol)
CMKOH = 0.4 / 0.2 = 2M
\(n_{K_2O} = \dfrac{9,4}{94}=0,1(mol)\\ K_2O + H_2O \to 2KOH\\ n_{KOH} = 2n_{K_2O} = 0,2(mol)\\ n_{KOH\ trong\ dd} = 0,2 + \dfrac{200.5,6\%}{56} = 0,4(mol)\\ C_{M_{KOH}} = \dfrac{0,4}{0,2} = 2M\)