The answer is 110., but round accordingly
Answer:
18 moles
Explanation:
Here the combustion of one mole of glucose ----> carbon dioxide + water, releases 2870 kilojoules / moles.
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With one contraction cycle requiring 55 kilojoules,
2870 / 55 ≈ 52.18
And with the efficiency being 35 percent,
52.1818..... * 0.35 = ( About ) 18 moles
<u><em>Hope that helps!</em></u>
I believe the end result is still 83 moles since there is never an amount of sulfur atoms added to the initial amount, but rather oxygen and water is repeatedly added to it. To find it's weight, first find the molar mass of H2SO4:
H2 + S + O4 = 2.00 + 32.1 + 64.0 = 98.1 g/mol
and mass = (98.1 g/mol)(83 mol) = 8142.3 g
rounded to 8.1 x 10^3 g assuming 100% yield?
2.4 mol NaSO2 × (6.02×10^23) f.units ÷ 1 mol
=1.44480000×10^25
Approximately 1.4×10^25, if you use sig.figs.
