Aj thjch gjaj baj tap hoa thj vao day cung gjaj nao moj nguoi.?

Question

Cho 1 anken A ket hop vs 1 hjdro duoc ankan B.
a) Xd CTPT of A va B.bjet rang de dot chay het B banh 1 luong oxi vua du thi the tich khi CO2 thu duoc bang 1 nua tong the tich cua B va oxi.
b) 1 hh X gom A,B va hjdro vs Vx = 22,4 l.cho X wa njken nung nong,xuc tac thu duoc hh Y voi ti khoi hoi cua X so voi Y la 0,7.tinh the tich cua Y , so mol hjdro va A da pu vs nhau.
c) bjet rang hh Y khong lam mat mau dd brom va ti khoi of Y so vs hjdro = 16.xd thanh phan % the tich cua X.(tat ca de o dktc).
baj nay mjnh moj gjaj ra cau a thuj ak.A la C3H6.B la C3H8.cac ban co the khoi gjaj cau a cung duoc.thanks.

? · Accepted Answer

A: CnH2n ---> B: CnH2n+2
CnH2n+2 + (3n+1)/2O2 ---> nCO2 + (n+1)H2O
1 --------------(3n+1)/2 ------------ n

n = 1/2[1 + (3n+1)/2] --> n = 3 --> C3H6 & C3H8

b) nX = 1 mol
dX/Y = 0.7 --> MX = 0.7MY <=> mX/nX = 0.7mY/nY
BTKL --> mX = mY --> 1/nX = 0.7/nY --> nY = 0.7nX = 0.7 --> VY = 15.68L

1 mol C3H6 pu' thi` tao thanh` 1 mol C3H8 tuong ung' nen Vgiam (so' mol giam) chinh' la` VH2pu' voi' C3H6 --> nH2pu' = nC3H6pu' = 1 - 0.7 = 0.3

c) Y ko lam` mat' mau` dd Br2 --> C3H6 het' --> nC3H6 = 0.3 --> %VC3H6 = 30%
X gom`: 0.3 mol C3H6, x mol C3H8, y mol H2 ---> x + y = 0.7 (1)
Y gom`: (x + 0.3) mol C3H8; (y - 0.3) mol H2
MY = [44(x + 0.3) + 2(y - 0.3)]/0.7 = 16*2 --> 44x + 2y = 9.8 (2)
(1) (2) ---> x = 0.2; y = 0.5 --> %VC3H8 = 20%; %VH2 = 50%

Aj thjch gjaj baj tap hoa thj vao day cung gjaj nao moj nguoi.?

1 Câu trả lời