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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Answer: Carbohydrates (carbo- = “carbon”; hydrate = “water”) contain the elements carbon, hydrogen, and oxygen, and only those elements with a few exceptions. The ratio of carbon to hydrogen to oxygen in carbohydrate molecules is 1:2:1.
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Answer:
6.31g/mol
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
Mole (n) = mass (m)/molar mass (Mm)
* Mm = m/n
Also, density (p) = mass (m) ÷ volume (V)
PV = nRT
Since n = M/Mm
PV = M/Mm. RT
PV × Mm = m × RT
Divide both sides by V
P × Mm = m/V × RT
Since p = m/V
P × Mm = p × RT
Mm = p × RT/P
Mm = 0.249 × 0.0821 × 293/0.95
Mm = 5.989 ÷ 0.95
Mm = 6.31g/mol