AgNO3 + NaCl → AgCl + NaNO3 được Trường TCSP Mẫu giáo – Nhà trẻ Hà Nội biên soạn gửi tới bạn đọc là phương trình phản ứng AgNO3 ra AgCl. Hy vọng tài liệu sẽ giúp các bạn viết và cân bằng đúng phương trình phản ứng, từ đó vận dụng tốt vào giải các dạng bài tập câu hỏi tương tự. Mời các bạn tham khảo.
1. Phương trình phản ứng AgNO3 tác dụng NaCl
AgNO3 + NaCl → AgCl↓ + NaNO3
2. Điều kiện phản ứng xảy ra
Nhiệt độ thường
3. Hiện tượng phản ứng giữa AgNO3 tác dụng NaCl
Khi cho dung dịch AgNO3 tác dụng với dung dịch NaCl, sau phản ứng tạo thành bạc clorua không tan có màu trắng
4. Bài tập vận dụng liên quan
Câu 1. Cho dung dịch AgNO3 vào dung dịch NaCl thì có hiện tượng là:
A. Có bọt khí thoát ra, dung dịch thu được không màu
B. Có xuất hiện kết tủa trắng
C. Dung dịch đổi màu vàng nâu
D. Không có hiện tượng gì xảy ra
Câu 2. Kết tủa hoàn toàn m gam NaCl bởi dung dịch AgNO3 dư thấy thu được 2,87 gam kết tủa. Giá trị của m là
A. 11,7 gam.
B. 1,71 gam.
C. 17,1 gam.
D. 1,17 gam.
nkết tủa = 2,87/143,5 = 0,02 mol
Phương trình phản ứng hóa học
AgNO3 + NaCl → AgCl↓ + NaNO3
0,02 ← 0,02
mNaCl = 0,2. 58,5 = 11,7 gam
Câu 3. Cho m gam muối NaCl tác dụng với dung dịch AgNO3 dư thì thu được một kết tủa , kết tủa này sau khi phản ứng phân hủy hoàn toàn cho 1,08 gam bạc. Tính giá trị của m
A. 0,585 gam
B. 5,850 gam
C. 1,17 gam
D. 1,755 gam
NaCl + AgNO3 → NaNO3 + AgCl
2AgCl 
2Ag + Cl2
nAg =1,08/108=0,01 (mol)
==> nAgCl = nAg = 0,01 (mol)
nNaCl = nAgCl = 0,01 (mol)
==> mNaCl = 0,01.58,5=0,585 (g)
Vậy m = 0,585 gam.
Câu 4. Cho phản ứng NaOH + HCl → NaCl + H2O. Phản ứng hóa học nào sau đây có cùng phương trình ion thu gọn với phản ứng trên?
A. 2KOH + FeCl2 → Fe(OH)2 + 2KCl.
B. NaOH + NH4Cl → NaCl + NH3 + H2O.
C. NaOH + HNO3 → NaNO3 + H2O.
D. AgNO3 + NaCl → AgCl↓ + NaNO3
Phương trình ion rút gọn của phản ứng NaOH + HCl → NaCl + H2O là:
OH– + H+ → H2O
A. 2OH– + Fe2+ → Fe(OH)2
B. OH– + NH4+ → NH3 + H2O
C. OH– + H+ → H2O
D. Ag+ + Cl– → AgCl
Câu 5. Trong các cặp chất cho dưới đây, cặp chất nào có thể cùng tồn tại trong một dung dịch?
A. AlCl3 và CuSO4.
B. NH3 và AgNO3.
C. AgNO3 và NaCl .
D. NaHSO4 và NaHCO3.
Câu 6. Cho một mẫu K vào 200ml dung dịch AlCl3 thu được 2,8 lit khí (đktc) và một kết tủa A. Nung A đến khối lượng không đổi thu được 2,55 gam chất rắn. Tính nồng độ mol/l của dung dịch AlCl3
A. 0,375M
B. 0,75M
C. 0,025M
D. 0,45M
mrắn: Al2O3 → nAl2O3 = 0,025 mol
→ nAl(OH)3 = 0,05 mol
nKOH = 2nH2 = 0,25 mol.
TH1: KOH thiếu, chỉ có phản ứng.
3KOH + AlCl3 → Al(OH)3 + 3KCl
Không xảy ra vì số mol Al(OH)3 tạo ra trong phản ứng > số mol Al(OH)3 đề cho.
TH2: KOH dư, có 2 phản ứng xảy ra.
3KOH + AlCl3 → Al(OH)3 + 3KCl
0,15 0,05 0,05 mol
4KOH + AlCl3 → KAlO2 + 3KCl + H2O
(0,25 – 0,15) 0,025
Tổng số mol AlCl3 phản ứng ở 2 phương trình là 0,075 mol
→ Nồng độ của AlCl3 = 0,375M
Câu 7. Cho m gam Fe tác dụng hết với dung dịch CuSO4 dư, thu được 28,8 gam Cu. Giá trị của m là
A. 50,4.
B. 12,6.
C. 16,8.
D. 25,2.
nCu = 0,45 mol
Phương trình hóa học
CuSO4 + Fe → FeSO4 + Cu
0,45 ← 0,45 mol
⟹ mFe = 0,45.56 = 25,2 gam
……………………………..
Mời các bạn tham khảo thêm tài liệu liên quan.
Trên đây Trường TCSP Mẫu giáo – Nhà trẻ Hà Nội vừa giới thiệu tới các bạn phương trình hóa học AgNO3 + NaCl → AgCl + NaNO3, mong rằng qua bài viết này các bạn có thể học tập tốt hơn môn Hóa lớp 12. Mời các bạn cùng tham khảo thêm các môn Ngữ văn 12, Tiếng Anh 12, Thi thpt Quốc gia môn Toán, Thi THPT Quốc gia môn Vật Lý,….
Ngoài ra, Trường TCSP Mẫu giáo – Nhà trẻ Hà Nội đã thành lập group chia sẻ tài liệu học tập THPT miễn phí trên Facebook: Tài Liệu Học Tập Trường TCSP Mẫu giáo – Nhà trẻ Hà Nội . Mời các bạn học sinh tham gia nhóm, để có thể nhận được những tài liệu mới nhất.
Chúc các bạn học tập tốt.
Bản quyền bài viết thuộc trường trung học phổ thông Sóc Trăng. Mọi hành vi sao chép đều là gian lận.
Nguồn chia sẻ: Trường THPT Thành Phố Sóc Trăng (thptsoctrang.edu.vn)
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