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belka [17]
3 years ago
13

When 10.0 g of sulfur is combined with 10.0 g of oxygen, 20.0 g of sulfur dioxide is formed. What mass of oxygen would be requir

ed to convert 10.0 g of sulfur into sulfur trioxide?
Chemistry
1 answer:
Illusion [34]3 years ago
3 0

Answer:

Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

6 g S * 1 mol/32.06 g S = 0.187 mol S

Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂

Since the molar mas of O₂ is 32 g/mol,

Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂

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x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

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Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF

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Assuming that a % x of the solute dissociates, we have the ICE table:

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initial concentration:       0.0854                    0             0

final concentration:      0.0854(1-x/100)   0.0854x/100   0.0854x / 100

We see that sum of molality of equilibrium mixture = freezing point molality

0.0854( 1 - x/100 + x/100 + x/100) = 0.187

2.1897 = 1 + x / 100

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