\(n_{KOH}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\)(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

=> m_H2SO4 = 0,02.98 = 1,96 (g)

=> \(m_{ddH_2SO_4}=\dfrac{1,96.100}{35}=5,6\left(g\right)\)

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

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

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

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

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

Đáp án: C

giải chi tiết:

\(n_{KOH}=\dfrac{11,2.20\%}{56}=0,04\left(mol\right)\)

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

           0,04mol    0,02mol

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

\(\Rightarrow m_{d_2H_2SO_4}\)\(_{35\%}\) = \(\dfrac{1,96}{35\%}=5,6\left(g\right)\)

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

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.

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

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

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

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)

\(\text{1)}m_{KOH}=40.35\%=14\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{14}{56}=0,25\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \text{Theo pthh}:n_{HCl}=n_{KOH}=0,25\left(mol\right)\\ \rightarrow V_{ddHCl}=0,25.0,5=0,125\left(l\right)\)

\(\text{2)}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ n_{H_2SO_4}=200.14,7\%=29,4\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\\ \text{PTHH}:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ \text{LTL}:\dfrac{0,15}{2}< \dfrac{0,3}{3}\rightarrow H_2SO_4\text{ dư}\)

\(\text{Theo pthh}:\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\n_{H_2}=n_{H_2SO_4\left(pư\right)}=0,225\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,15=0,075\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{dd\left(\text{sau phản ứng}\right)}=200+4,05-0,3.2=203,45\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\text{ dư}}=\dfrac{\left(0,3-0,225\right).98}{203,45}=3,61\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,075}{203,45}=12,61\%\end{matrix}\right.\)

Chọn B