Question

Bài 2. Trung hòa 200g ddNaOH 10% bằng x(g) ddH2SO4 15%  dd muối y%. Tính x, y.

Answer 1

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{NaOH}=\dfrac{200\cdot10\%}{40}=0,5\left(mol\right)\) \(\Rightarrow n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x=m_{ddH_2SO_4}=\dfrac{0,25\cdot98}{15\%}\approx163,3\left(g\right)\\m_{Na_2SO_4}=0,25\cdot142=35,5\left(g\right)\end{matrix}\right.\)

\(\Rightarrow y=C\%_{Na_2SO_4}=\dfrac{35,5}{163,3+200}\cdot100\%\approx9,77\%\)