Question

Để trung hòa 11,2 gam KOH 20% thì cần lấy bao nhiêu gam dung dịch axit H₂SO₄ 35%

Answer 1

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

\(PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ \: ­\: ­\: ­\: ­\: ­\: ­\: ­\: \: \: ­\: ­\: ­\: ­\: ­\: ­0,04mol\: ­\: ­\: ­\: ­\: 0,02mol\)

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

\(\Rightarrow m_{d^2\:H_2SO_4\:35\%}=\dfrac{1,96}{35\%}=5,6g\)

Answer 2

Ta có n_KOH = \(\dfrac{11,2\times20\%}{56}\) = 0,04 ( mol )

2KOH + H₂SO₄ \(\rightarrow\) K₂SO₄ + 2H₂O

0,04..........0,02

=> m_H2SO4 = 0,02 . 98 = 1,96 ( gam )

=> m_{H2SO4 cần dùng} = 1,96 : 35 . 100 = 5,6 ( gam )