Asked by tomi
The combustion of hydrogen–oxygen mixtures is used to produce very high temperatures (approximately 2500 °C) needed for certain types of welding operations. Consider the reaction to be
H2(g) + 1/2O2(g) --> H20(l)
Standard enthalpy is -241.8kj
What is the quantity of heat evolved,
in kilojoules, when a 140g mixture containing equal parts of H2 and O2 by mass is burned?
H2(g) + 1/2O2(g) --> H20(l)
Standard enthalpy is -241.8kj
What is the quantity of heat evolved,
in kilojoules, when a 140g mixture containing equal parts of H2 and O2 by mass is burned?
Answers
Answered by
DrBob222
Please don't change your screen name when posting different questions. It gets us confused.
This is a limiting reagent problem as well as a thermodynamic one. First, determine the limiting reagent. I didn't work through all of the problem but I think the limiting reagent is oxygen. Check me out on that. Calculate how many mols H2O are formed from the limiting reagent. The reaction produces 241.8 kJ/1 mol H2O so mols x kJ/mol should give you the answer. (I divided the 140 gram mixture into 70 g hydrogen and 70 g oxygen since the problem said equal parts by mass.)
This is a limiting reagent problem as well as a thermodynamic one. First, determine the limiting reagent. I didn't work through all of the problem but I think the limiting reagent is oxygen. Check me out on that. Calculate how many mols H2O are formed from the limiting reagent. The reaction produces 241.8 kJ/1 mol H2O so mols x kJ/mol should give you the answer. (I divided the 140 gram mixture into 70 g hydrogen and 70 g oxygen since the problem said equal parts by mass.)
Answered by
Shashi Naraine
yaya that's right, lol assuming u neeed it for mastering chemistry, but i don't get why you need to use the moles of H20.
Answered by
tpain
you don't have to, you can just use the moles of the limiting reagent and multiply it by 241.8kJ/mol
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