Химия

Уравнение реакции:

$Ba (OH)_{2} + H_{2}SO_{4} = BaSO_{4} + 2H_{2}O$.

в $m_{1}$ г раствора $Ba (OH)_{2}$ находится $m_{1} \frac {c_{1}}{100}$ г, или $\frac {m_{1}c_{1}}{17100}$ моль основания. В растворе $H_{2}SO_{4}$ находится $m_{2} \frac {c_{2}}{100}$ г или $\frac {m_{2}c_{2}}{9800}$ моль кислоты. Если $\frac {m_{1}c_{1}}{17100}$ \frac {M (H_{2}SO_{4}) \left ( \frac {m_{2}c_{2}}{9800} - \frac {m_{1}c_{1}}{17100} \right ) 100}{m_{1} + m_{2} - m (BaSO_{4})} = \frac {m_{2}c_{2} - \frac{98}{171} m_{1}c_{1}}{m_{1} + m_{2} - \frac {233m_{1}c_{1}}{17100}}$(%)$H_{2}SO_{4}$. Если$ \frac {m_{1}c_{1}}{17100} > \frac {m_{2}c_{2}}{9800}$, то в растворе останется$Ba (OH)_{2}$:$ \frac {M (Ba (OH)_{2}) \left ( \frac {m_{1}c_{1}}{17100} - \frac {m_{2}c_{2}}{9800} \right )}{m_{1} + m_{2} - m(BaSO_{4})}100 = \frac {m_{1}c_{1} - \frac {171}{98m_{2}c_{2}}}{m_{1} + m_{2} - \left ( \frac {233m_{2}c_{2}}{9800} \right )}$%$Ba (OH)_{2}$.$x$ л