Question

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

Answer 1

Ta có: \(C_{\%_{KOH}}=\dfrac{m_{KOH}}{11,2}.100\%=20\%\)

=> m_KOH = 2,24(g)

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

PTHH: 2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{KOH}=\dfrac{1}{2}.0,04=0,02\left(mol\right)\)

=> \(m_{H_2SO_4}=0,02.98=1,96\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{1,96}{m_{dd_{H_2SO_4}}}.100\%=35\%\)

=> \(m_{dd_{H_2SO_4}}=5,6\left(g\right)\)