During the combustion of organic matter with a volume of 10 liters, 40 liters were formed. carbon dioxide and 40.18 g of water. The vapor of this substance is twice as heavy as nitrogen. Find the molecular formula for this hydrocarbon.
V (CxHy) = 10L;
V (CO2) = 40l, m (H2O) = 40.18g
DN2 = 2
1. Reaction equation:
CxHy + О2 = хСО2 + y / 2H2O.
2. According to the equation, from 10 liters of CxHy, 40 liters of CO2 are formed, and from 22.4 liters of CxHy – 22.4 x liters of CO2. Let’s make the proportion and find x.
10 / 22.4 = 40 / 22.4x, x = 40/10 = 4 (C)
3. According to the equation, 40.18 g of H2O is formed from 10 liters of CxHy, and from 22.4 liters – y / 2 * 18g (ie 9y). Let’s make the proportion and find y.
10 / 22.4 = 40.18 / 9y, y = 22.4 * 40.18 / 90 = 10 (H).
4. Find Мr (CxHy) and Мr (CxHy).
Mr (CxHy) = Mr (N2) * 2 = 28 * 2 = 56. Mr (C4H10) = 58, which means that the true formula is C4H8.
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