Lời giải:· \(BTKL:{m_Y} + {m_A} = {m_B} + {m_Z} + {m_{{H_2}O}} \to {m_{{H_2}O}} = 2,52\left( {gam} \right) \to {n_{{H_2}O}} = 0,14\left( {mol} \right)\)
· \(BT{\rm{ }}H:{n_{KHS{O_4}}} = 4{n_{NH_4^ + }} + 2{n_{{H_2}O}} \to {n_{NH_4^ + }} = \frac{{0,36 - 2.0,14}}{4} = 0,02\left( {mol} \right)\)
· \(BT{\rm{ }}N:{n_{N\left( Z \right)}} = {n_{KN{O_3}}} - {n_{NH_4^ + }} = 0,02\left( {mol} \right);{m_Z} = {m_{N\left( Z \right)}} + {m_{O\left( Z \right)}} = 0,6\left( {gam} \right)\)
\( \to {m_{O\left( Z \right)}} = 0,32\left( {gam} \right) \to {n_{O\left( Z \right)}} = 0,02\left( {mol} \right)\)
· \(BT{\rm{ }}O:{n_{O\left( Y \right)}} + 3{n_{KN{O_3}}} = {n_{O\left( Z \right)}} + {n_{{H_2}O}} \to {n_{O\left( Y \right)}} = 0,04\left( {mol} \right)\)
· \({m_{KL\left( X \right)}} = {m_Y} - {m_{O\left( Y \right)}} = 6,17 - 0,04.16 = 5,53\left( {gam} \right)\) (1)
Do Cu chiếm 34,72% về khối lượng trong X \(\to {m_{Cu}} = 1,92\left( {gam} \right) \to {n_{Cu}} = 0,03\left( {mol} \right)\)
\(BTe:{n_{e\left( {KL} \right)}} = 2{n_{SO_4^{2 - }}} - {n_{{K^ + }}} - {n_{NH_4^ + }} = 2.0,36 - 0,4 - 0,02 = 0,3\left( {mol} \right)\) (2)
Từ (1), (2): \(\left\{ \begin{array}{l}
56a + 27b + \underbrace {1,92}_{{m_{Ca}}} = 5,53\\
3a + 3b + \underbrace {0,03.2}_{{n_{e\left( {Cu} \right)}}} = 0,3
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,05 = {n_{Fe}}\\
b = 0,03 = {n_{Al}}
\end{array} \right.\)
B + NaOH → lượng OH- dùng để kết hợp các cation \(\left( {C{u^{2 + }},F{e^{3 + }},A{l^{3 + }},NH_4^ + } \right)\) là:
\({n_{O{H^ - }}} = \underbrace {0,03.3}_{{n_{A{l^{3 + }}}}} + \underbrace {0,05.3}_{{n_{F{e^{3 + }}}}} + \underbrace {0,03.2}_{{n_{C{u^{2 + }}}}} + \underbrace {0,02}_{{n_{NH_4^ + }}} = 0,32\left( {mol} \right) \to {n_{O{H^ - }{\rm{do }}}} = \underbrace {0,34}_{{\rm{ban dau}}} - \underbrace {0,32}_{\left( {{\rm{pu}}} \right)} = 0,02\left( {mol} \right)\)
0,02 mol OH- dư hòa tan \(Al{\left( {OH} \right)_3} \to Al\left( {OH} \right)_4^ - \to {n_{Al{{\left( {OH} \right)}_3}{\rm{ du}}}} = 0,03 - 0,02 = 0,01\left( {mol} \right)\)
\(\Rightarrow {m_{{\rm{ket tua}}}} = \underbrace {{m_{Fe{{\left( {OH} \right)}_3}}}}_{0,05.107} + \underbrace {{m_{Cu{{\left( {OH} \right)}_2}}}}_{0,03.98} + \underbrace {{m_{Al{{\left( {OH} \right)}_{3{\rm{ }}\left( {{\rm{du }}} \right)}}}}}_{0,01.78} = 9,07\left( {gam} \right)\).
