Giáo án hóa học 8 trọn bộ chuẩn

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Tiết 19: PHẢN ỨNG HÓA HỌC (tiếp) I. MỤC TIÊU 1. Kiến thức (Hs nắm được) Để nhận biết có phản ứng hóa học xảy ra, dựa vào một số dấu hịệu có chất mới tạo thành mà ta quan sát được như thay đổi màu sắc, tạo kết tủa, khí thoát ra … 2. Kỹ năng Nhận biết dấu hiệu của phản ứng Viết phương trình chữ của phản ứng 3. Thái độ Nghiêm túc học tập, cẩn thận khi làm thí nghiệm II. TRỌNG TÂM Dấu hiệu nhận biết có phản ứng xảy ra

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Khối lượng mỗi nguyên tố: mH = 5 ,88 34 = 2(g), mS = 34 – 2 = 32(g) 100 mH = + Số mol nguyên tử mỗi nguyên tố: nH = 2 32 = 2(mol), nS = = 1(mol), 1 32 5 ,88 34 = 2(g), 100 mS = 34 – 2 = 32(g) + Số mol nguyên tử mỗi nguyên + CTHH cần tìm là H2S Giáo án Hóa Học 8 Bài 5 trang 71 + Khối lượng mol chất A: MA = 17.2 = 34(g) + Khối lượng mỗi nguyên tố: 78 Năm học 2013 - 2014 Trường THCS... trong đá vôi: %CaCO3 = 250 100% = 89 ,3% 280 CaCO3 100% %CaCO3 = m đa voi m = 250 100% = 89 ,3% 280 - Hs làm bài tập trên bảng, Hs ở dưới nhận xét, sửa sai (nếu có) - Gv nhận xét, nêu đáp án BT4/61/SGK a) C2H4 + 3O2 → 2CO2 + 2H2O b) Số pt C2H4 : số pt O2 = 1 : 3 Số pt C2H4 : số pt CO2 = 1 : 2 BT5/61/SGK a) Tìm chỉ số x, y : Giáo án Hóa Học 8 57 Năm học 2013 - 2014 Trường THCS... sao biết khối lượng “mol không khí” là 29 gam? - Hs: (trả lời) - Gv: không khí gồm 78% khí N2, 21% khí O2 và 1% là các khí khác, người ta xem như không khí gồm 80 % N2 và 20% là O2 nên trong một mol không khí có 0 ,8 mol N2 và 0,2 mol O2 Khối lượng “mol không khí” là: Mkk = (28g x 0 ,8) + (32g x 0,2) ≈ 29 g Tóm lại: muốn biết khí A nặng hay nhẹ hơn không không bao nhiêu... b Đề kiểm tra A Phần trắc nghiệm Câu 1: 0,125 mol O2 có khối lượng là: a 4 gam b 8 gam c 16 gam d 24 gam Câu 2: 8 gam O2 tương ứng với số mol là: a 0,125 mol b 0,25 mol c 0,5 mol d 0,75 mol Câu 3: 0,375 mol khí nitơ ở đktc có thể tích là: a 2 ,8 lít b 5,6 lít c 8, 4 lít d 11,2 lít Câu 4: 16 ,8 lít khí nitơ ở đktc tương ứng với số mol là: a 0,15 mol b 0,25 mol c 0,5 mol d 0,75... làm vào vở, nhận xét bài làm trên bảng 28 64 = 0,5(mol), nCu = = 1(mol) 56 64 5,4 nAl = = 0,2 (mol) 27 BT3/67/sgk 28 64 = 0,5(mol), nCu = = 1(mol) 56 64 5,4 nAl = = 0,2 (mol) 27 a) nFe = a) nFe = b) VCO2 = 0,175.22,4 = 3,92 (lít) V H2 = 1,25.22,4 = 28 (lít) V N2 = 3.22,4 = 67,2 (lít) V b) CO2 = 0,175.22,4 = 3,92 (lít) V H2 = 1,25.22,4 = 28 (lít) V N2 = 3.22,4 = 67,2 (lít) - Gv nhận xét,... THCS Lê Lợi Huy Gv Mai Đức = 129,552 (lít) - Gv hướng dẫn: đổi khối lượng các khí ra mol - Hs thực hiện: 1 3,5 n = 0,5 N2= = 0,125 2 28 8 33 n O2 = = 0,25 nCO2 = = 0,75 32 44 n H2 = BT6/67/sgk 1 = 0,5 2 8 n O2 = = 0,25 32 n H2 = 3,5 = 0,125 28 33 n CO2 = = 0,75 44 n N2 = - Gv: Thể tích chất khí tỷ lệ thuận với số mol nên ta vẽ theo số mol + Vẽ một khối bất kỳ tương ứng... lượng các nguyên tố có trong các hợp chất Fe2O3 và SO3 Giải + MFe2O3 = 56.2 + 16.3 = 160 %Fe = %O = 100% - 70% = 30% + MSO3 = 32 + 16.3 = 80 112 100% = 70% 160 %O = 100% - 70% = 30% + SO3 = 32 + 16.3 = 80 M 32 %S = 100% = 40% 80 %S = 32 100% = 40% 80 %O = 100% - 40% = 60% - Gv nhận xét %O = 100% - 40% = 60% 5 Hướng dẫn – Dặn do - Làm bài tập 1 trang 71 SGK vào vở bài tập - Tìm hiểu... (g) 100 100 mO = 106 – (46 + 12) = 48 (g) + Số mol nguyên tử mỗi nguyên tố: *Luyện tập Bài 2b trang 71 + Khối lượng mỗi nguyên tố: 43,4 106 = 46(g), 100 11,3 mC = 106 = 12 (g) 100 mNa = mO = 106 – (46 + 12) = 48 (g) + Số mol nguyên tử mỗi nguyên tố: 46 12 = 2(mol), nC = = 1(mol), 23 12 48 nO = = 3 (mol) 16 nNa = 46 = 2(mol), 23 12 nC = = 1(mol), 12 48 nO = = 3 (mol) 16 nNa = + CTHH cần . C*?+6D9 E F)GH%=I#C*H@/I#/*+FC@#F  JK L #  JK L #MC@  #KF E AN N(,&#5OP#0QR#(S  E T";UV B FJ;&*"1- 67189:; <'=! *. ! "U 345 B D9*?+r& E F)`tK L #/*+  CK  #CHKFI  #F  K E AN N(,&# /0(,&# 5OP#0QR# (6# 0h6. 9N0+@*"7/ 345 B C*?+6D9 E F)*+  JK L #MC@  E AN N(6&@#_,( E T";UV B FJezY

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