0.210g 得碳氫化合物完全燃燒後產生0.660g之CO2,問其實實驗為何?

乂乂熼乂乂

2008-03-24 3:35 am

( 一) 0.210g 得碳氫化合物完全燃燒後產生0.660g之CO2,問其實實驗為何?
若於S.T.P.狀態下,測得其密度為1.87g/dm3,則分子式為何?

(二) 2.86g的1-丁烯( C4H10 )和丁烷 C4H10與過量的氧混合燃燒產生 8.80gCO2 及 4.14g水,求丁烷在原來混合物中所佔質量百分率.

(提示: 設X mol C4H8; Y mol C4H10, 然後利用 C 原子及 H 原子的 mol數不變的方法,列放程式,解 X 及 Y之值)

C= 12 H= 1 O= 16

超級緊急的!!!

回答 (1)

用戶7369

2008-03-24 5:57 am

✔ 最佳答案

1. 2CxHy+(2x+y/2)O2------->2xCO2+yH2O
the number of mole of hydrocarbon=0.21(12x+y)
the number of mole of carbon dioxide formed=0.66/44
the ratio is 1:x
after solving, 2x=y
therefore, C:H=1:2
the density =1.87
for one mole of gas, the mass =22.4x1.87=42
12x+y=42
14x=42
x=3
y=6
the formula is C3H6
2. C4H8 +6O2------->4CO2 + 4H2O
C4H10+13/2------->4CO2+5H2O
let x be the mass of C4H10
the no. of mole of CO2 formed =8.8/44
(2.86-x)/56x4+x/58x4=8.8/44
after solving x=1.74
the %by mass of C4H10=1.74/2.86x100%
60.8%

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