Question

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Answer 1

$CuO + 2 HCl \to {CuCl}_{2} + H_{2} O x 2 x {Fe}_{2} O_{3} + 6 HCl \to 2 {FeCl}_{3} + 3 H_{2} O y 6 y \{\begin{array}{l} 80 x + 160 y = 19, 2 \\ 2 x + 6 y = 0, 2 x 3, 4 = 0, 68 \end{array} \Rightarrow \{\begin{cases} x = 0, 04 \\ y = 0, 1 \end{cases} % CuO = 0, 04 x 80 / 19, 2 x 100 % = 16, 67 % % {Fe}_{2} O_{3} = 83, 33 % b, {nCuCl}_{2} = 0, 04 mol \Rightarrow CM {CuCl}_{2} = 0, 04 / 0, 2 = 0, 2 M {nFeCl}_{3} = 0, 1 mol \Rightarrow CM {FeCl}_{3} = 0, 1 / 0, 2 = 0, 5 M$

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Answer 2

$CuO + 2 HCl \to {CuCl}_{2} + H_{2} O x 2 x {Fe}_{2} O_{3} + 6 HCl \to 2 {FeCl}_{3} + 3 H_{2} O y 6 y \{\begin{array}{l} 80 x + 160 y = 19, 2 \\ 2 x + 6 y = 0, 2 x 3, 4 = 0, 68 \end{array} \Rightarrow \{\begin{cases} x = 0, 04 \\ y = 0, 1 \end{cases} % CuO = 0, 04 x 80 / 19, 2 x 100 % = 16, 67 % % {Fe}_{2} O_{3} = 83, 33 % b, {nCuCl}_{2} = 0, 04 mol \Rightarrow CM {CuCl}_{2} = 0, 04 / 0, 2 = 0, 2 M {nFeCl}_{3} = 0, 1 mol \Rightarrow CM {FeCl}_{3} = 0, 1 / 0, 2 = 0, 5 M$

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Answer 3

$n_{HCl} = 0, 2.3, 4 = 0, 68 mol CuO + 2 HCl - > {CuCl}_{2} + H_{2} O x 2 x x (mol) {Fe}_{2} O_{3} + 6 HCl - > 2 {FeCl}_{3} + 3 H_{2} O y 6 y 2 y (mol) Ta có : \{\begin{array}{l} 2 x + 6 y = 0, 68 \\ 80 x + 160 y = 19, 2 \end{array} < = > \{\begin{cases} x = 0, 04 \\ y = 0, 1 \end{cases} a, mCuO = 0, 04.80 = 3, 2 g = > % m_{CuO} = 16, 67 % % m_{{Fe}_{2} O_{3}} = 100 % - 16, 67 % = 83, 33 % b, n_{{FeCl}_{3}} = 2 y = 0, 2 mol = > C_{M {FeCl}_{3}} = 0, 2 : 0, 2 = 1 M n_{{CuCl}_{2}} = x = 0, 04 mol = > C_{M {CuCl}_{2}} = 0, 04 : 0, 2 = 0, 2 M$ .

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Answer 4

$n_{HCl} = 0, 2.3, 4 = 0, 68 mol CuO + 2 HCl - > {CuCl}_{2} + H_{2} O x 2 x x (mol) {Fe}_{2} O_{3} + 6 HCl - > 2 {FeCl}_{3} + 3 H_{2} O y 6 y 2 y (mol) Ta có : \{\begin{array}{l} 2 x + 6 y = 0, 68 \\ 80 x + 160 y = 19, 2 \end{array} < = > \{\begin{cases} x = 0, 04 \\ y = 0, 1 \end{cases} a, mCuO = 0, 04.80 = 3, 2 g = > % m_{CuO} = 16, 67 % % m_{{Fe}_{2} O_{3}} = 100 % - 16, 67 % = 83, 33 % b, n_{{FeCl}_{3}} = 2 y = 0, 2 mol = > C_{M {FeCl}_{3}} = 0, 2 : 0, 2 = 1 M n_{{CuCl}_{2}} = x = 0, 04 mol = > C_{M {CuCl}_{2}} = 0, 04 : 0, 2 = 0, 2 M$ .

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