Question

Câu 1. Để trung hòa 11,2gam KOH 20%, thì cần lấy bao nhiêu gam dung dịch axit H₂SO₄35%

               a. 9gam                          b. 4,6gam             c. 5,6gam                    d. 1,7gam

Answer 1

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

⇒

Answer 2

C

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

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

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

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

 

Answer 3

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

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

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

Answer 4

5,6