Đáp án: C

giải chi tiết:

$n_{KOH}=\frac{11,2.20\%}{56}=0,04mol$

$PTHH:2KOH+H_{2}S{O}_4\to K_{2}S{O}_4+2H_{2}O­­­­­­­­­­­­­0,04mol­­­­0,02mol$

$\Rightarrow {}^{}$

C

\(n_{KOH}=\dfrac{11,2.20\%}{56}=0.04mol\)

\(PTHH:2KOH+H_2SO_4->K_2SO_4+2H_2O\)

\(=>m_{H_2SO_4}=98.0,02=1,96\left(g\right)\)

\(=>m_{d^2H_2SO_4}35\%=\dfrac{1.96}{35\%}=5,6g\)

\(n_{OH^-}=n_{KOH}=\dfrac{1,12}{56}=0,02mol\)

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

Chọn B

Câu 1 : 

\(n_{KOH}=\dfrac{25\%.112}{100\%.56}=0,5\left(mol\right)\)

Pt : \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

\(n_{H2SO4}=\dfrac{1}{2}n_{KOH}=0,25\left(mol\right)\Rightarrow m_{ddH2SO4}=\dfrac{0,25.98}{4,9\%}.100\%=500\left(g\right)\)

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; \(C_{\%_{HCl}}=\dfrac{1,825}{m_{dd_{HCl}}}.100\%=20\%\)

=> \(m_{dd_{HCl}}=9,125\left(g\right)\)

cảm ơn nhìu

\(m_{KOH}=\dfrac{11,2\cdot20\%}{100\%}=2,24\left(g\right)\\ \Rightarrow n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=0,02\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,02\cdot98=1,96\left(g\right)\\ \Rightarrow A\)

Đáp Án: D. 7,84g

Ta có: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

 ⇒ \(m_{KOH}=2,24\) (mol)

⇒ \(n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\)

⇒ \(n_{H_2SO_4}=2n_{KOH}=0,08\left(mol\right)\)

⇒ \(m_{H_2SO_4}=0,08.98=7,84\left(g\right)\)

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)