calculate volume of air required to burn 5.8 gram of C4 h10 at ntp
Answers
Alright, your chemical equation is not balanced. It should be 2 C4H10 + 13O2 →10 H2O + 8 CO2
This originally caught me off-guard and messed up my calculations giving a result about twice what is actually the volume needed for complete combustion.
Alright, so you got 50 cm^3 of butane at standard temperature and pressure. According to Wolfram Alpha, the density of butane at 0 degrees Celsius and atmospheric pressure (101.325 kPa) is 0.002704 g/cm^3
0.002704 times 50 is 0.1352 grams. So we got 0.1352 grams of butane. The molar mass of butane is 58.12 g/mol.
Dividing 0.1352/58.12 gives us 0.00232622161 moles of butane. Now, a complete combustion would require 6.5 moles of dioxygen (O2) for every mole of butane. So, 0.00232622161 times 6.5 gives us 0.0151204405 moles of O2. The molar mass of dioxygen is 31.9988 g/mol, give or take. 31.9988 g/mol times 0.0151204405 moles gives us a total of 0.483835951 grams of O2.
The density of Oxygen at STP is 0.001429 g/cm^3 . So, dividing 0.483835951 grams by 0.001429 g/cm^3 gives us 338.583591 cm^3.
338.583591 cm^3 of oxygen are needed to completely burn 50 cm^3 of butane. The byproducts are irrelevant.
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