Tài liệu 500 bài tập luyện thi ptth quốc gia và đại học môn hóa 2016 (ôn tổng lực 2 tháng cuối) ( www.sites.google.com/site/thuvientailieuvip )

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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