Propane & propane accessories are used in state of the art grilling equipment. Propane gas (C3H8) is burned in air to generate the heat needed in outdoor grilling. How many grams of oxygen are required to react with 59.5 g of propane?
Answer: 216 g
Someone please explain. Studying for my final tomorrow!
Three answers:
Chris
2013-04-29 13:52:30 UTC
This is straightforward stoichiometry. There are many good tutorials on the web, and especially on YouTube.
BB
2013-04-30 00:25:08 UTC
C3H8 + 5O2 -----> 3 CO2 + 4 H2O + heat.
From the stoichiometric coefficients we know
that 5 moles O2 react with 1 mole C3H8.
Moles Propane = mass / molar mass =
59.5 g / 44.0 g/mole = 1.352.
Moles O2 = 1.352 * 5 = 6.76.
Mass O2 = mol * molar mass
= 6.76 mol * 32 g/mol = 216 g.
randhir
2016-08-08 01:16:42 UTC
The ultimate fuel law does not remember molecular interactions. The answer is 1 on account that if lots of the area is empty, the molecules will not be shut enough to have interaction.
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