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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,
From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with
L of oxygen gas.
So, 0.720 L of ammonia will react with:
Therefore, the volume of oxygen required is 900 mL.
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14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.
Explanation:
The balanced equation for the reaction is to be known so that number of moles taking part can be known.
The balanced chemical equation is
4Fe + 3⇒ 2
From the given weight of iron to be used for the production of , number of moles of Fe taking part in the reaction can be known by the formula:
Number of moles= mass ÷ Atomic mass of one mole of the element.
(Atomic weight of Fe is 55.845 gm/mole)
Putting the values in equation
Number of moles = 10 gm ÷ 55.845 gm/mole
= 0.179 moles
Applying the stoichiometry concept
4 moles of Fe gives 2 Moles of Fe2O3
0.179 moles will produce x moles of Fe2O3
So, 2÷ 4 = x ÷ 0.179
2/4 = x/ 0.179
2 × 0.179 = 4x
2 × 0.179 / 4 = x
x = 0.0895 moles
So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.
Now the formula used above will give the weight of Fe2O3
weight = atomic weight × number of moles
= 159.69 grams × 0.0895
= 14.292 grams of Fe2O3 formed.