Ngâm 22,6 g hỗn hợp ba kim loại Zn, Fe, Cu trong dung dịch H2SO4 loãng. Phản ứng xong thu được 3g chất rắn không tan và 6,72 lít khí ở đktc a/ Viết các pt xảy ra b/ Phần trăm khối lượng mỗi kim loại trong hỗn hợp Hứa vote đủ

chipper2609

2 năm trước

Loga Sinh Học lớp 12

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Đáp án:

$\begin{array}{l}
\% Cu = 13,9\% \\
\% Zn = 60,2\% \\
\% Fe = 25,9\%
\end{array}$

Giải thích các bước giải:

21,6g hôn hợp kim loại nha em

$\begin{array}{l}
a)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{m_{Cu}} = 3g\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Zn(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
65a + 56b = 18,6
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{Zn}} = 0,2 \times 65 = 13g\\
\% Cu = \dfrac{3}{{21,6}} \times 100\% = 13,9\% \\
\% Zn = \dfrac{{13}}{{21,6}} \times 100\% = 60,2\% \\
\% Fe = 100 - 13,9 - 60,2 = 25,9\%
\end{array}$

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vurtung.300103

(1)(2)$\to$ hệ nghiệm âm

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