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?
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.)
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.
you don't have to, you can just use the moles of the limiting reagent and multiply it by 241.8kJ/mol
To calculate the quantity of heat evolved when a mixture containing equal parts of H2 and O2 is burned, we need to use the stoichiometry of the reaction and the standard enthalpy change.
First, let's determine the moles of hydrogen and oxygen in the mixture:
1. Calculate the moles of hydrogen (H2):
Mass of H2 = 140g * (1/2) = 70g
Moles of H2 = Mass of H2 / Molar mass of H2
The molar mass of hydrogen (H2) is 2g/mol. So,
Moles of H2 = 70g / 2g/mol = 35 mol H2
2. Calculate the moles of oxygen (O2):
Since the mixture contains equal parts of H2 and O2 by mass, the moles of O2 will be the same as H2.
Therefore, Moles of O2 = 35 mol O2
Next, we need to use the stoichiometry of the reaction to determine the heat evolved:
From the balanced equation, we can see that the reaction consumes 1 mol of H2 and produces 1 mol of H2O.
The standard enthalpy change for the reaction is -241.8 kJ/mol, which means 1 mol of H2 reacts to produce 241.8 kJ of heat.
To calculate the heat evolved for 35 moles of H2, we can use the following proportion:
35 moles H2 / 1 mol H2 = x kJ / 241.8 kJ
Simplifying the proportion:
35 moles H2 * 241.8 kJ = x kJ
x = (35 mol H2) * (241.8 kJ/mol)
Calculating the heat evolved:
x = 8463 kJ
Therefore, when a 140g mixture containing equal parts of H2 and O2 by mass is burned, the quantity of heat evolved is approximately 8463 kJ.