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Đổi 100ml = 0,1 lít
Ta có: \(C_{M_{KOH}}=\dfrac{n_{KOH}}{0,1}=0,5M\)
=> \(n_{KOH}=0,05\left(mol\right)\)
a. PTHH: 2KOH + H2SO4 ---> K2SO4 + 2H2O
b. Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{KOH}=\dfrac{1}{2}.0,05=0,025\left(mol\right)\)
Vì H2SO4 là chất lỏng nên thể tích bằng số mol của chính nó.
c. PTHH: KOH + HCl ---> KCl + H2O
Theo PT: \(n_{HCl}=n_{KOH}=0,05\left(mol\right)\)
=> \(m_{HCl}=0,05.36,5=1,825\left(g\right)\)
Ta có; \(C_{\%_{HCl}}=\dfrac{1,825}{m_{dd_{HCl}}}.100\%=20\%\)
=> \(m_{dd_{HCl}}=9,125\left(g\right)\)
\(n_{KOH}=0,04.2=0,08\left(mol\right)\)
PTHH: 2KOH + CuSO4 --> Cu(OH)2\(\downarrow\) + K2SO4
______0,08------------------->0,04
=> mCu(OH)2 = 0,04.98 = 3,92(g)
=> A
\(n_{KOH}=2.0,04=0,08\left(mol\right)\)
PTHH: 2KOH + CuSO4 --> Cu(OH)2↓↓ + K2SO4
______0,08------------------->0,04
=> mCu(OH)2 = 0,04.98 = 3,92(g)
=> A
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)
K2O + H2O -> 2KOH (1)
nK2O=\(\dfrac{14,1}{94}=0,15\left(mol\right)\)
Theo PTHH 1 ta có:
2nK2O=nKOH=0,3(mol)
CM dd KOH=\(\dfrac{0,3}{0,5}=0,6M\)
c;
2NaOH + H2SO4 -> Na2SO4 + 2H2O (2)
Theo PTHH 2 ta có:
\(\dfrac{1}{2}\)nNaOH=nH2SO4=0,15(mol)
mH2SO4=98.0,15=14,7(g)
mdd H2SO4=14,7:10%=147(g)
Vdd H2SO4=147:1,14=129(ml)
a, K2O + H2O ->2 KOH
b, nK2O= 0,15 ( mol )
K2O + H2O-> 2KOH
Theo pt 1 1 2 ( mol )
Theo đb 0,15 0,15 0,3 ( mol)
==> CM=\(\dfrac{0.3}{0,5}\) = 0,6 M
c, 2KOH + H2SO4 -> K2SO4 + 2H2O
Theo pt 2 1
Theo đb 0,15 0,075
==> mdd H2S04=\(\dfrac{0,075.98.100\%}{10\%}=73,5\left(g\right)\)
==> Vdd h2so4=\(\dfrac{73,5}{1,14}=64,47\left(ml\right)\)
\(n_{KOH}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\)(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)
=> \(m_{ddH_2SO_4}=\dfrac{1,96.100}{35}=5,6\left(g\right)\)
C
\(n_{KOH}=\dfrac{11,2.20\%}{56}=0.04mol\)
\(PTHH:2KOH+H_2SO_4->K_2SO_4+2H_2O\)
\(=>m_{H_2SO_4}=98.0,02=1,96\left(g\right)\)
\(=>m_{d^2H_2SO_4}35\%=\dfrac{1.96}{35\%}=5,6g\)
Trung hoà 20ml dd H2SO4 1M bằng dd NaOH 20%.
\(n_{H_2SO_4}=0,02mol\)
H2SO4+2NaOH\(\rightarrow\)Na2SO4+2H2O
\(n_{NaOH}=2n_{H_2SO_4}=0,04mol\)
\(m_{NaOH}=0,04.40=1,6gam\)
H2SO4+2KOH\(\rightarrow\)K2SO4+2H2O
\(n_{KOH}=2n_{H_2SO_4}=0,04mol\)
\(m_{KOH}=0,04.56=2,24gam\)
\(m_{dd_{KOH}}=\dfrac{2,24.100}{5,6}=40gam\)
\(V_{KOH}=\dfrac{40}{1,045}\approx38,3ml\)
\(m_{KOH}=\dfrac{11,2\cdot20\%}{100\%}=2,24\left(g\right)\\ \Rightarrow n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=0,02\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,02\cdot98=1,96\left(g\right)\\ \Rightarrow A\)
Đáp Án: D. 7,84g
Ta có: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
⇒ \(m_{KOH}=2,24\) (mol)
⇒ \(n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
⇒ \(n_{H_2SO_4}=2n_{KOH}=0,08\left(mol\right)\)
⇒ \(m_{H_2SO_4}=0,08.98=7,84\left(g\right)\)