How many moles of nitrogen are in 73.0g of nitrous oxide n2o

Hey there!:

Molar Mass

N2O = 44.013 g/mol

Therefore:

number of moles N :

73.0 g * 1 mol N2O / 44.013 g N2O * 2 mols N / 1 mol N2O

73.0 * 1 / 44.013 * 2 / 1 =

73.0 / 44.013 * 2 =

1.6586 * 2 => 3,31 moles of N

hope this helps!

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Alleei Alleei · Answer 2 · 2019-09-30T12:35:08+02:00

Answer : The number of moles of nitrogen present in nitrous oxide is 3.32 moles.

Explanation : Given,

Mass of nitrous oxide = 73.0 g

Molar mass of nitrous oxide = 44 g/mole

Now we have to calculate the moles of [tex]N_2O[/tex].

Formula used :

[tex]\text{ Moles of }N_2O=\frac{\text{ Mass of }N_2O}{\text{ Molar mass of }N_2O}[/tex]

[tex]\text{ Moles of }N_2O=\frac{73.0g}{44g/mole}=1.66moles[/tex]

Now we have to calculate the moles of nitrogen in nitrous oxide.

In [tex]N_2O[/tex] molecule, there are 2 moles of nitrogen atoms and 1 mole of oxygen atom.

As, 1 mole of [tex]N_2O[/tex] contains 2 moles of nitrogen

So, 1.66 moles of [tex]N_2O[/tex] contains [tex]1.66\times 2=3.32[/tex] moles of nitrogen.

Therefore, the number of moles of nitrogen present in nitrous oxide is 3.32 moles.