A reduction in the mass of oxygen available to react with the hydrogen,
reduces the heat produced.
Reasons:
The chemical equation for the reaction obtained online is presented as
follows;
[tex]\displaystyle H_2(g) + \frac{1}{2} O_2 (g) \longrightarrow H_2O(g), \ \Delta H^{\circ} = -285.82 \, kJ[/tex]
From the above equation, one mole of H₂ reacts with half [tex]\left(\dfrac{1}{2} \right)[/tex] mole of O₂
to produce 285.82 kJ.
Therefore;
[tex]\displaystyle 55.0 \, kJ \ will \ be \ produced \ by \ \dfrac{55.0}{285.22} \times\dfrac{1}{2} = \frac{1375}{14291} \approx 9.62 \times 10^{-2} \ moles \ of O_2[/tex]
Molar mass of O₂ = 31.999 g/mol
Mass = Number of moles × Molar mass
Mass of oxygen gas, O₂, required, m, is therefore;
m = 9.62 × 10⁻² moles × 31.999 g/mol ≈ 3.08 grams
Learn more here:
https://brainly.com/question/14796831
The equation in HQ 11.43 as obtained online is presented as follows;
[tex]\displaystyle H_2(g) + \frac{1}{2} O_2 (g) \longrightarrow H_2O(g), \ \Delta H^{\circ} = -285.82 \, kJ[/tex]
