How many grams of oxygen gas are needed to react completely with 44.2 grams iron

When discussing how many grams of oxygen gas are needed to react completely with 44.2 grams iron, it is important to understand the chemical equation for the reaction and apply stoichiometric principles. The equation for this reaction is as follows:

Fe + O₂ → Fe₃O₄

In order to determine the amount of oxygen gas (in grams) needed to react completely with 44.2 g of iron, we must use stoichiometry principles. In this case, we need to calculate the moles of both reactants in order to determine how much oxygen (in moles) will be necessary for a complete reaction.

How many grams of oxygen gas are needed to react completely with 44.2 grams iron

Using Avogadro’s number (6 x 10²³ mol⁻¹), we can begin by determining the moles of iron present in 44.2 g:
Moles = Mass/Molar Mass = 44.2g/55.845g/mol = 0.791 mol Iron

We can also use Avogadro’s number (6 x 10²³ mol⁻¹) along with molar mass data from periodic table elements in order to determine the total amount of oxygen required for a complete reaction:
Moles = Mass/Molar Mass = 32g/15.999 g/mol = 2mol Oxygen

We know that during a balanced chemical equation each side must possess an equal number of atoms or molecules; therefore, if 0.791 mole Iron reacts completely with 2mole Oxygen then 0.791 mole Oxygen will remain unused after completion of the reaction instead 2mole – 0 . 79 1mole= 1 . 21mole remaining unused . So now we have all information required which means that :
GramsOxygen= Molesx MolarMass = 1 . 21 mol x 15 . 999 g / mol = 19 . 2459 gram Oxygen gas

Therefore, 19.2459grams Oxygen Gas is required to react completely with 44.2grams Iron according to given equation “Fe+ O₂ → Fe₃O₄”

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