Naczynie I: [latex]\M_{CaO}=56frac{g}{mol}\m=8g\n=frac{m}{M}=frac{8g}{56frac{g}{mol}}approx0,143mol\\ M_{H_{2}O}=18frac{g}{mol}\m=100g\n=frac{m}{M}=frac{100g}{18frac{g}{mol}}approx5,556mol[/latex] ............CaO..................+........H₂O.........⇒...............Ca(OH)₂............. ____________ __________ _____________ 1mol : 1mol : 1mol 0.143mol : 5.556mol : 0.143mol niedomiar przereaguje w całości nadmiar przereaguje 0.143mol pozostanie: 5.556-0.143mol=5.413mol [latex]\n_{s}=0.143mol\ M_{Ca(OH)_{2}}=74frac{g}{mol}\ m_{s}=0.143mol*74frac{g}{mol}=10.582g\\ n_{H_{2}O}=5.413mol\ M_{H_{2}O}=18frac{g}{mol}\ m_{H_{2}O}=5.413mol*18frac{g}{mol}=97.434g\\\\ m_{s}=10.582g\ m_{r}=10.582g+97.434g=108.016g\ C_{p}=frac{m_{s}*100\%}{m_{r}}=frac{10.582g*100\%}{108.016g}approx9,8\%[/latex] Naczynie II: [latex]\m_{s}=8g\ m_{r}=100g+8g=108g\\ C_{p}=frac{m_{s}*100\%}{m_{r}}=frac{8g*100\%}{108g}approx7.407\%[/latex] Pozdrawiam :)
