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Trang chủ Giáo dục - Đào tạo Luyện thi - Đề thi Thi THPT Quốc Gia Hóa học 500 bài tập luyện thi ptth quốc gia và đại học môn hóa 2016...

Tài liệu 500 bài tập luyện thi ptth quốc gia và đại học môn hóa 2016

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www.facebook.com/groups/TaiLieuOnThiDaiHoc01 oc 01 FC – HOÁ HỌC VÙNG CAO 2.0 ---Ad:DongHuuLee--- D ai H CHIẾN LƯỢC ÔN TẬP TỔNG LỰC PHÂN TÍCH – GIỚI THIỆU PHƯƠNG PHÁP TƯ DUY VÀ CÁC KĨ THUẬT GIẢI NHANH w w w .fa ce bo o k. co m /g ro up s/ Ta iL ie uO nT hi 500 BÀI TẬP HOÁ HỌC www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Tác giả : Lê Hữu Đông – THPT Cẩm Thủy I – Cẩm Thủy – Thanh Hóa facebook : DongHuuLee PHẦN 1 .CÁC KĨ THUẬT TƯ DUY & GIẢI CÁC BÀI TẬP HOÁ HỌC TRONG ĐỀ THI TUYỂN SINH ĐẠI HỌC – CAO ĐẲNG . oc 01 Bài 1: Cho 19,3 gam hỗn hợp bột Zn và Cu có tỉ lệ mol tương ứng là 1 : 2 vào dung dịch chứa 0,2 mol Fe2(SO4)3. Sau khi các phản ứng xảy ra hoàn toàn, thu được m gam kim loại. Giá trị của m là A. 6,40 B. 16,53 C. 12,00 D. 12,80 ( Trích Câu 2- Mã đề 596 – ĐH khối A – 2010) ro up s/ Ta iL ie uO nT hi D ai H Cần biết • Khi cho nhiều kim loại tác dụng với muối thì phản ứng xảy ra theo quy tắc α và kim loại mạnh phản ứng trước, kim loại yếu phản ứng sau. • Khi viết phản ứng giữa kim loại không tan trong nước với muối ta nên viết phản ứng ở dạng ion. • Các kim loại từ Fe đến Cu đều có khả năng kéo Fe3+  → Fe2+. M + nFe3+  → Mn+ + nFe2+ • Các kim loại không tan trong nước, trước Fe có thể kéo Fe3+ về Fe: Ban đầu,kim loại M kéo Fe3+ về Fe2+: M + nFe3+  → Mn+ + nFe2+ 2+ Sau đó nếu M còn thì M sẽ kéo Fe về Fe : 2M+ nFe2+  → 2Mn+ + nFe ( Lượng Fe2+ bị kéo về Fe một phần hay toàn bộ là phụ thuộc vào lượng kim loại M còn nhiều hay ít) • Khi giải bài tập dạng này có thể dùng hai cách: - Cách1: Phương pháp 3 dòng ( tức tính tại ba thời điểm : ban đầu, phản ứng, sau phản ứng ). - Cách 2: Dựa vào ĐL bảo toàn electron kinh nghiệm: i Trong mọi phản ứng, kim loại luôn cho e và : ne cho = nkl.hoá trị i Ion kim loại thường nhận e và : ne nhận = độ giảm số O ion kl .nion kl Vậy: m /g nkl . hoá trị = độ giảm số O ion kl .nion nB = y.nAx By ce bo o k. co (Trong giới hạn của đề thi,hầu hết các ion kim loại đều đóng vai trò là chất nhận e và số e nhận thường bằng điện tích của ion .Riêng các ion Fe2+, Fe+8/3, Cu+ và Cr2+ là có khả năng cho e và khả năng này chỉ được bộc lộ khi chúng tác dụng với O2, HNO3 và H2SO4 đặc). • Cách tính số mol của nguyên tố trong phân tử: Với hợp chất AxBy thì: nA = x.nAx By w w w .fa Công thức này có vẻ đơn giản nhưng hiệu quả của nó thì không hề nhỏ.Thời gian sẽ cho các em biết điều đó !!! Tóm tắt: Bài giải Zn  ddFe2 ( SO4 )3 :0,2 mol 19,3g hh  → m (g) Kim loại = ? Cu Cách 1. Phương pháp 3 dòng Ta có: nZn bđ = 0,1mol; nCu bđ = 0,2mol, nFe3+ = 0,2.2 = 0,4mol www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 w w w .fa ce bo o k. co m /g ro up s/ Ta iL ie uO nT hi D ai H oc 01 012456789  40 0000 274 8 ! 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