Answer:
1360kJ are evolved
Explanation:
When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.
To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:
Moles H2 -Molar mass: 2g/mol-
90g H2 * (1mol / 2g) = 45 moles
Moles O2 -Molar mass: 32g/mol-
90g * (1mol / 32g) = 2.81moles
For a complete reaction of 2.81 moles of O2 are needed:
2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2
As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.
As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:
2.81 moles O2 * (241.8kJ / 1/2moles O2) =
The quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is
The equation for the combustion of hydrogen is given as:
[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]
Recall that:
Since the mass of 180 g is equally shared by H₂ and O₂, then:
The number of moles of the reactant can be determined as follows:
For H₂:
[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]
no of moles = 44.6 mol
For O₂:
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]
no of moles = 2.8 mol
Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:
If 1/2 moles of O₂ produces -241.8 kJ/mol of water;
Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:
[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]
[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]
= - 1358.91 kJ
≅ - 1360 kJ
Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ
Learn more about the quantity of heat here:
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