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\(n_{KOH}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\)(mol)
PTHH : 2KOH + H2SO4 ---> K2SO4 + 2H2O
2 : 1 : 1 : 2
0.2mol 0.1mol
\(m_{H_2SO_4}=n.M=0,1.98=9,8\left(g\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{9,8.100\%}{35\%}=28\left(g\right)\)
\(n_{KOH}=\dfrac{11,2.20}{100.56}=0,04\left(mol\right)\)
PTHH: 2KOH + H2SO4 --> K2SO4 + 2H2O
_____0,04---->0,02
=> mH2SO4 = 0,02.98 = 1,96 (g)
=> \(m_{ddH_2SO_4}=\dfrac{1,96.100}{35}=5,6\left(g\right)\)
Ta có: \(C_{\%_{KOH}}=\dfrac{m_{KOH}}{11,2}.100\%=20\%\)
=> mKOH = 2,24(g)
=> \(n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
PTHH: 2KOH + H2SO4 ---> K2SO4 + 2H2O
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{KOH}=\dfrac{1}{2}.0,04=0,02\left(mol\right)\)
=> \(m_{H_2SO_4}=0,02.98=1,96\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{1,96}{m_{dd_{H_2SO_4}}}.100\%=35\%\)
=> \(m_{dd_{H_2SO_4}}=5,6\left(g\right)\)
\(n_{OH^-}=n_{KOH}=\dfrac{1,12}{56}=0,02mol\)
Để trung hòa\(\Rightarrow n_{H^+}=n_{OH^-}=0,02mol\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{H^+}=0,01mol\Rightarrow m=0,98g\)
\(m_{dd}=\dfrac{0,98}{35}\cdot100=2,8\left(g\right)\)
Chọn C.
có thể cho em hỏi chỗ oh- với oh+ là sao 0 ạ?? em 0 hiểu chỗ đó lắm
Câu 1 :
\(n_{KOH}=\dfrac{25\%.112}{100\%.56}=0,5\left(mol\right)\)
Pt : \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(n_{H2SO4}=\dfrac{1}{2}n_{KOH}=0,25\left(mol\right)\Rightarrow m_{ddH2SO4}=\dfrac{0,25.98}{4,9\%}.100\%=500\left(g\right)\)
\(m_{KOH}=\dfrac{11,2\cdot20\%}{100\%}=2,24\left(g\right)\\ \Rightarrow n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=0,02\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,02\cdot98=1,96\left(g\right)\\ \Rightarrow A\)
Đáp Án: D. 7,84g
Ta có: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
⇒ \(m_{KOH}=2,24\) (mol)
⇒ \(n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
⇒ \(n_{H_2SO_4}=2n_{KOH}=0,08\left(mol\right)\)
⇒ \(m_{H_2SO_4}=0,08.98=7,84\left(g\right)\)
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
nH2SO4=0,02.1=0,02(ol)
a) PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
0,04____________0,02____0,02(mol)
mNaOH=0,04.40= 1,6(g)
=>mddNaOH= (1,6.100)/20= 8(g)
b) PTHH: H2SO4 + 2 KOH -> K2SO4 + 2 H2O
0,2____________0,04(mol)
=>mKOH=0,04.56=2,24(g)
=>mddKOH= (2,24.100)/5,6=40(g)
=>VddKOH= mddKOH/DddKOH= 40/1,045=38,278(ml)
Đáp án: C
giải chi tiết:
nKOH=11,2.20%56=0,04molnKOH=11,2.20%56=0,04mol
PTHH:2KOH+H2SO4→K2SO4+2H2O0,04mol0,02molPTHH:2KOH+H2SO4→K2SO4+2H2O0,04mol0,02mol
⇒
C
\(n_{KOH}=\dfrac{11,2.20\%}{56}=0.04mol\)
\(PTHH:2KOH+H_2SO_4->K_2SO_4+2H_2O\)
\(=>m_{H_2SO_4}=98.0,02=1,96\left(g\right)\)
\(=>m_{d^2H_2SO_4}35\%=\dfrac{1.96}{35\%}=5,6g\)