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How many moles of oxygen gas are needed to react completely with 17.9 mL Mg, with a density of 1.74 g/mL?


unbalanced equation: Mg + O2 “yields”/ MgO


0.212 mol O2


0.641 mol O2


1.28 mol O2


2.56 mol O2

Verified Answer :

MissPhiladelphia

For the answer to the question above, 17.9 x 1.74 gram Mg = 31.15 gram Mg = (31.15/24.31) = 1.28 mole Mg 

2 Mg + O2 ---> 2MgO 

so 1.28 mole Mg reacts with 0.641 mole O2

I hope my answer helped you. Feel free to ask more questions. Have a nice day!

Answer:

0.641 O2

Explanation:

Volume of Mg = 17.9 ml

Density of Mg = 1.74 g/ml

Mass of Mg = \(density*volume = 1.74 g/ml*17.9 ml = 31.15 g\)

Atomic mass of Mg = 24 g/mol

Moles of Mg present = \(\frac{Mass}{atomic mass} = \frac{31.15}{24} = 1.298 moles\)

The balanced reaction is:

2Mg + O2\rightarrow 2MgO

Based on the reaction stoichiometry:

2 moles of Mg reacts with 1 mole of O2

Moles of O2 that would combine with 1.298 moles of Mg is:

= \(\frac{1.298\ Mg*1\ O2}{2\ Mg}= 0.649 moles\)

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