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How many moles of oxygen are needed to react with 6.78 grams of ammonia?

4NH3 + 5O2-> 4NO + 6 H2O

Respuesta :

Answer: 0.499mol

Explanation:

4NH3 + 5O2 -> 4NO + 6H2O

Molar Mass of NH3 = 14 + 3 = 17g/mol

Mass conc. Of NH3 = 6.78g

n = Mass conc. / molar Mass

n = 6.78 / 17 = 0.399mol

From the equation,

4moles of NH3 required 5moles of O2.

Therefore, 0.399mol of NH3 will require Xmole of O2. i.e

Xmol of O2 = (0.399 x 5) / 4 = 0.499mol

The number of moles of oxygen required is 0.499mol.

Given,

Molar Mass of [tex]NH_3 = 14 + 3 = 17g/mol[/tex]

Mass concentration of [tex]NH_3=6.78 g[/tex]

The formula for number of moles :

[tex]n=\frac{Given mass}{Molar mass}[/tex]

We can calculate the moles by using the above formula,

n = 6.78 / 17 = 0.399mol

From the equation,

4moles of [tex]NH_3[/tex] required 5moles of [tex]O_2[/tex].

Therefore, 0.399mol of [tex]NH_3[/tex] required x moles of [tex]O_2[/tex].

X mol of [tex]O_2 = (0.399\times 5) / 4 = 0.499mol[/tex]

Find more information about number of moles here,

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