When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:

The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
More on stoichiometric calculations can be found here; brainly.com/question/27287858
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Explanation:
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<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>
In general, the gas equation can be written

where
P = pressure, atm , N/m²
V = volume, liter
n = number of moles
R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)
T = temperature, Kelvin
n = N / No
n = mole
No = Avogadro number (6.02.10²³)
n = m / MW
m = mass
MW = molecular weight
For density , can be formulated :

P = 327 mmHg = 0,430263 atm
R = 0.082 L.atm / mol K
T = 48 ºC = 321.15 K
MW of H₂ = 2.015 g/mol
The density :

Answer:
96.32 %
Explanation:
Given that:
The solubility of compound in hot water = 4.35 g / 100 mL
The solubility of compound in cold water = 0.16 g / 100 mL
Which means that in 100 mL of hot water, the dissolved compound is 4.35 g and in cold water, the dissolved compound is 0.16 g
Hence, on transition, compound that will catalyze is 4.35 - 0.16 g = 4.19 g
So,

Percent recovery for re-crystallization of this compound from water= 96.32 %