Bước 3: Xác định chất dư sau phản ứng Số mol \(HCl\) ban đầu là:\(n_{HCl}=C_{M}\times V=3,5\times 0,05=0,175(mol)\)So sánh tỉ lệ số mol theo phương trình:\(\frac{n_{Fe}}{1}=\frac{0,05}{1}=0,05\)\(\frac{n_{HCl}}{2}=\frac{0,175}{2}=0,0875\)Vì \(0,0875>0,05\) nên \(HCl\) là chất còn dư sau phản ứng. 

Bước 2: Tính thể tích khí hydrogen (đkc) Theo phương trình, tỉ lệ mol giữa \(Fe\) và \(H_{2}\) là \(1:1\), nên:\(n_{H_{2}}=n_{Fe}=0,05(mol)\)Thể tích khí \(H_{2}\) ở điều kiện chuẩn (đkc ở \(25^{\circ }C\), \(1bar\)) là:\(V_{H_{2}}=0,05\times 24,79=1,2395(L)\)

Bước 1: Viết phương trình hóa học và tính số mol Fe Số mol của \(Fe\) tham gia phản ứng là:\(n_{Fe}=\frac{2,8}{56}=0,05(mol)\)Phương trình hóa học:\(Fe+2HCl\rightarrow FeCl_{2}+H_{2}\)Bước 2: Tính thể tích khí hydrogen (đkc) Theo phương trình, tỉ lệ mol giữa \(Fe\) và \(H_{2}\) là \(1:1\), nên:\(n_{H_{2}}=n_{Fe}=0,05(mol)\)Thể tích khí \(H_{2}\) ở điều kiện chuẩn (đkc ở \(25^{\circ }C\), \(1bar\)) là:\(V_{H_{2}}=0,05\times 24,79=1,2395(L)\)Bước 3: Xác định chất dư sau phản ứng Số mol \(HCl\) ban đầu là:\(n_{HCl}=C_{M}\times V=3,5\times 0,05=0,175(mol)\)So sánh tỉ lệ số mol theo phương trình:\(\frac{n_{Fe}}{1}=\frac{0,05}{1}=0,05\)\(\frac{n_{HCl}}{2}=\frac{0,175}{2}=0,0875\)Vì \(0,0875>0,05\) nên \(HCl\) là chất còn dư sau phản ứng. Đáp án: a. Phương trình: \(Fe+2HCl\rightarrow FeCl_{2}+H_{2}\)b. Thể tích khí \(H_{2}\): 1,2395 Lc. Chất dư: Dung dịch HCl

Rút gọn các biểu thức sau: a) √ 36 x 8 ( 2 − y ) 2 36 𝑥 8 ( 2 − 𝑦 ) 2 với y ≥ 2 𝑦 ≥ 2 ; b) √ 7 z 3 . √ 3 28 z 7 𝑧 3 . 3 28 𝑧 với z > 0 𝑧 > 0 .

Rút gọn các biểu thức sau: a) √ 36 x 8 ( 2 − y ) 2 36 𝑥 8 ( 2 − 𝑦 ) 2 với y ≥ 2 𝑦 ≥ 2 ; b) √ 7 z 3 . √ 3 28 z 7 𝑧 3 . 3 28 𝑧 với z > 0 𝑧 > 0 .

Một hình chữ nhật có chiều dài là √ a 3 𝑎 3 mét và chiều rộng là √ a 12 𝑎 12 (mét) ( a > 0 ) ( 𝑎 > 0 ) . Tính diện tích của hình chữ nhật theo a.

We'll see you again in September.

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