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2 years ago

What volume of o2 is needed to react fully with 720. Ml of nh3

1 answer:

6 0

Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.

<h3>What is volume?</h3>

Volume is the area occupied by the substance and is the ratio of the mass to the density.

At STP, 1 mole of gas occupies 22.4 L of volume

Given,

Volume of ammonia reacted = 0.720 L

The combustion reaction is shown as,

$\rm 4NH_{3} + 5O_{2} \rightarrow 4NO +6H_{2}O$

From the stoichiometry of the reaction, it can be said that,

$(4 \times 22.4)$ L of ammonia reacts with $(5 \times 22.4)$ L of oxygen gas.

So, 0.720 L of ammonia will react with:

$\dfrac{ (5 \times 22.4)}{(4 \times 22.4)} \times 0.720 = 0.9 \;\rm L$

Therefore, the volume of oxygen required is 900 mL.

Learn more about volume here:

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Explanation:

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$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

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For the given chemical equation:

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By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitrogen dioxide

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Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

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