Bai 1:

 Goi ankin la C_nH_2n-2 a mol, C₂H₄ b mol.

Co a+b = 0,45, (pt do mol) (14n-2)a + 28b=12,1 (pt khoi luong), an +2b=0,9(pt so mol CO₂)

ó a=0,25 mol, b=0,2 mol, n=2. ó X la C₂H₂. Co m H₂O = 18(0,25+0,4)=11,7g.

Bai 2

Co n C₃H₄ = n ket tua =14,7/(108+12.3+3)=0,1 mol.

Co n hon hop khi ban dau la 0,4 mol.  ó n metan + n X=0,3 mol.

Dot chay metan va X dc n CO₂ =50/100 = 0,5 mol.

Co m CO₂ + mH₂O = 50 – 15,4 ó 0,5.44 + mH₂O = 34,6ó mH₂O=12,6g.

ó n H₂O=0,7 mol.

Co n metan = nH₂O – n CO₂ (do X dot chay tao ra so mol CO₂ va H₂O bang nhau) = 07-0,5=0,2 mol. ó nX = 0,1 mol. co n CO₂ do X tao ra= 0,5-0,2=0,3mol.

ó X la C₃H₆

C_nH_2n  +  3n/2 O₂   =  CO₂  +  H₂O

7                   25,2

ó 25,2 . 3n/2 = 7 ó n=2,4 .

do 2 olefin la the khi o dk thuong ó C₂H₄, C₃H₆, C₄H₈.

Nhung do n trung binh la 2,4 ó 2 olefin la C₂H₄ va C₃H₆ hoac C₂H₄ va C₄H₈

C   +   O₂   =  CO₂                           

a           a mol                                   

S   +   O₂   =   SO₂           (do khong co V₂O₅ nen phan ung dung lai o vc tao ra SO₂)

b         b mol

co a+b=0,5 va 12a + 32b =10 ó a=0,3 , b=0,2 mol.

ó mC= 3,6g, m S= 6,4g, V khi = 0,5. 22,4 = 11,2l

                 Gia su hon hop ban dau co 10 ml C₃H₈ va 10 ml O₂     

                          C₃H₈  +   5O₂   = 3CO₂   +   4H₂O

Ban dau   :          10         10               0           0 ml

Phan ung:           2            10              6           8 ml

Con lai  :              8            0               6            8 ml

Co V₀ = 10 + 10 = 20 ml, V’ = 8+6 =14 ml

ó V’ = 7/10 V₀

Co m H₂O = 4,32g ó n H₂O = 0,24 mol.

Co n O₂ = 0,24 mol. ó n O = 0,24.2 = 0,48 mol.

Co n O = 2.nCO₂ + nH₂O ó 0,48 = 2.nCO₂ + 0,24 ó nCO₂ = 0,12 mol.

Do X la hiđrocacbon va n CO₂ < nH₂O ó X la ankan.

ó n ankan = 0,24 – 0,12 =0,12 mol. ma nCO₂ =0,12 mol. ó X la CH₄

C_nH_2n   +   HOH   =   C_nH_2n+2O (Y)

C_nH_2n   +     HBr    =   C_nH_2n+1Br (Z)

Co m anken = 3,36g

Co cu 1 mol anken phan ung thi mZ- mY = 80 + 1 -2 -16=63g

     3,36/14n mol anken phan ung thi mZ - mY = 7,56g

ó7,56 = 63 * 3,36/14n ó n=2 ó anken la C₂H₄

Do 2 axit phan ung hoan toan nen n H₂ sinh ra = 1,8.0,05 + 1,2.0,05/2 =0,12 mol.

CuO + H₂ = Cu + H₂O

0,12                 0,12 mol

ó CuO du: 0,2-0,12 = 0,08 mol.

CuO + H₂SO₄ = CuSO₄ + H₂O

0,08 mol

Cu + 2H₂SO₄ = CuSO₄ + SO₂ + 2H₂O (vi day la H₂SO₄ dac)

0,12  0,24 mol

ó n H₂SO₄ can dung = 0,08 + 0,24 =0,32 mol. ó m H₂SO₄ =31,36g

ó m dung dich H₂SO₄ = 31,36*100/96=32,666g. ó V = 32,6666/1,84=17,7ml

Cho X tac dung vs H₂SO₄ loang thi co:

2Al + 3H₂SO₄=Al2(SO₄)3 + 3H₂

a                                                        1,5a mol

Mg + H₂SO₄= MgSO₄ + H₂

b                                         b mol

co 1,5a+b=0,5 mol.(1)

Cho X tac dung vs H₂SO₄  dac duoc:

Ap dung bao toan e co:

Al = Al⁺³  +  3e                             S⁺⁶   +   2e  = S⁺⁴

a                    3a                                               1,2      0,6mol

Mg = Mg⁺²   +   2e

  b                         2b

Cu= Cu⁺²   +     2e

c                             2c

ó 3a + 2b + 2c = 1,2.(2)

Do hon hop co khoi luong la 16,6g ó 27a + 24b+64c=16,6(3). Tu 1, 2, 3 ta duoc a=0,2 mol, b=0,2, c=0,1 mol. ó m Al=5,4g, mMg=4,8g, m Cu=6,4g

Gia su X la KL tac dung H₂SO₄ loang, Y la KL tac dung vs H₂SO₄ dac

X   +   H₂SO₄   =     XSO₄      +    H₂

0,1 mol

Ta co X= 6,5/0,1 = 65 ó X la Zn.

Y +   2H₂SO₄   =  YSO₄ +  SO₂ +2H₂O

                                               0,0025 mol

 Co m Y = 6,66 – 6,5 =0,16g ó Y=0,16/0,0025=64 ó Y la Cu