## Ai đó giúp em với ạ

Question

Ai đó giúp em với ạ

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7 months 2021-02-19T10:55:06+00:00 2 Answers 7 views 0

$$\begin{array}{l} Mg + 2HCl \to MgC{l_2} + {H_2}(1)\\ MgO + 2HCl \to MgC{l_2} + {H_2}O(2)\\ NaOH + HCl \to NaCl + {H_2}O(*)\\ {m_{HCl}} = \dfrac{{400 \times 7,3\% }}{{100\% }} = 29,2g\\ \to {n_{HCl}} = 0,8mol\\ {m_{NaOH}} = \dfrac{{100 \times 4\% }}{{100\% }} = 4g\\ \to {n_{NaOH}} = 0,1mol\\ \to {n_{HCl}}(*) = {n_{NaOH}} = 0,1mol\\ \to {n_{HCl}}(1) + (2) = 0,7mol\\ \left\{ \begin{array}{l} 24a + 40b = 11,6\\ 2a + 2b = 0,7 \end{array} \right. \to \left\{ \begin{array}{l} a = 0,15\\ b = 0,2 \end{array} \right.\\ \to {n_{Mg}} = 0,15mol\\ \to {n_{MgO}} = 0,2mol\\ a)\\ \% {m_{Mg}} = \dfrac{{0,15 \times 24}}{{11,6}} \times 100\% = 31,03\% \\ \% {m_{MgO}} = 100\% – 31,03\% = 68,97\% \\ b)\\ {n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,35mol\\ \to {m_{MgC{l_2}}} = 33,25g\\ {n_{{H_2}}} = {n_{Mg}} = 0,15mol \to {m_{{H_2}}} = 0,3g\\ {m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} – {m_{{H_2}}} = 411,3g\\ \to C{\% _{MgC{l_2}}} = \dfrac{{33,25}}{{411,3}} \times 100\% = 8,08\% \end{array}$$