C o6 Ig ma 't nhan dung cac dung dich khSngmau la: Na2S04 (1);
11FDun s6i dung dich khdng tháy c6 them Ret tiia chiing to trong dung did
Dun s6i dung dich khdng tháy c6 them Ret tiia chiing to trong dung did
kh6ng CO mu6i CăHC03)2. Vay chi c6 phan ung:
Ca (OH)^ + CO, ^ CaCO, + H^O (1)
S6' mol CaCOj = s6' mol CO2 = 0,55 mol ^ kh6'i luong CO2 = 24,2 (gamj So mol C = 0,55 => kh6'i luong C = 12.0,55 = 6,6 (gam).
^ kh6'i lucfng H = 7,6 - 6,6 = 1 (gam).
=^ S6 mol H = 1 => s6 mol H 2 O = 0,5. Phuong trinh hoa hoc:
b, ' b a + — 4. 3 x - l a C O ,+ - H 2 O C,H2,_2+ 2 0 2 - > x C 0 2 + ( x - l ) H 2 0 (2) (3) Tir (2) va (3) => 0,1a+ 0,05x = 0,55 (I) ^ 0,05b+ 0 , 0 5 ( x - l ) = 0,5 (II) Giai (I) va (II): 2a = b.
Cong thurc hidrocacbon A c6 th^ viet la C,H2^.
Co 2 trucmg hop xay ra, khi 2 hidrocacbon khac nhau 1 nguyen tir C.
+ Trucmg hap I: C.H^, va C,^^,)H,,^^,,_2
=0 0,1a+ 0,05(a + l ) - 0 , 5 5 Giaira a a 3,33 => loai!
+ Tnfdng hap 2: C^Hj, va C^,_,)ii2(.-i)-2
^ 0,1a+ 0,05(a-l) = 0,55
Giai ra a = 4 => CTPT hai hidrocacbon la C4HX va C3H4.
2. S6 mol CăOH)2 phan ung = 0,55 (mol). ^ S6' mol CăOH)2 du = — - 0,55 = 0,45 (mol).
74
=> Khoi luong Că0H)2 du = 0,45.74 = 33,3 (gam).
Khd'i luong dung dich sau khi loc két tua = m^^ + mc„, + m^ „ -mc^co,
= 1000 + 24,2 + 9-55 = 978,2 (gam
Sau khi loc két tiia, n6ng do % dung dich CăOH)2 = 33,3.100%
978,2 = 3,4'%
ill
Be so 5
(Th6i gian lam bai 150 phut) Cau 1 (2,5 diem)
Co 3 binh mat nhan, m6i binh chira m6t h6n hop dung djch sau: Na2C03 va K2SO4; NaHCO, va K 2 C O 3 ; NaHCO, va K2SO4. Trinh bay phuong phap hoa hoc d^ nhan biét 3 binh nay ma chi c^n diing them dung dich HCI va dung
djch BăN03)2 lam thu6'c thir. Cau 2 (2 diem)
Co can 160 gam dung dich CUSO4 10% dén khi tdng s6' nguyen tir trong dung dich chi con m6t n^a so v6i ban dSu thi dirng laị
Tim khd'i luong H2O bay rạ
Cau 3 (5,5 diem)
Nung a gam Cu trong b gam O2 thu dugc san pham X. X tan hoan toan
trong c gam dung djch H2SO4 n6ng dp 85% (dun nong) thu dugc dung dich Y
va khi Z. Toan b6 khi Z phan ung hét vdi oxi du (c6 xiic tac) tao ra oxit T, tat ca luong oxit nay phan ung vilra du v6i 150 ml dung djch NaOH 0,2M tao ra
2,62 gam muoi P. Cho dung djch Y tac dung vdi NaOH thi phai diing hét it
nhát 300 ml dung djch NaOH nong do I M mdi tao dugc lugng két tiia toi dạ
Neu cho dung djch Y tac dung vdi Bă0H)2 thi phai diing het it nhát v ml dung djch Bă0H)2 n6ng đ 1,5M mdi tao ra dugc lugng két tua td'i da la
44,75 gam. Tim a, b, c, v.
Cau4(2diém) ' Hay tim khd'i lugng nguyen tir ciia clo, kali, bac chi dua vao khd'i lugng
nguyen tir cua oxi cho 6 cud'i bai va qua trinh thi nghiem neu sau day:
- Nung 100 gam K C I O 3 (khan) thu dugc 39,17 gam oxi va 60,83 gam kali cloruạ
- Cho 100 gam kali clorua phan ling hoan toan vdi dung djch bac nitrat thu dugc 192,25 gam két tiiạ . ; , ,
- Phan tich 132,86 gam bac clorua tháy trong do cd 100 gam bac. Cau 5 (4 diem)
1. Dot chay trong oxi cac hidrocacbon cd cdng thiic la C„H2„^2,C„H2„,
C„H, n * 2 n - 2 -
a) Viet phuong trinh hoa hgc cua phan ling. b) Neu đ't 1 mol mdi chat, hay so sanh ti le: + So mol H 2 O : sd' mol COj.
+ Sd' mol C O 2: s6 mol Ộ
2. Co h6n hop khf A g6m CjHfi va CjHj. Cho biet hien tucfng xay ra