Answer
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Hint: Here you have to find the limiting reagent first. Predict the reaction according to the limiting reagent and you will get the reactant which will be left in excess.
Complete step-by-step answer:
Let us first write the complete reaction with the given reactants and conditions in mind. It is as follows:
\[2Mg(s)+{{O}_{2}}(g)\to 2MgO(s)\]
Here two moles of magnesium reacts with one mole oxygen to form two moles of magnesium oxide. We have to find out which reactant will be left in excess after the reaction is complete. For this we have to first find the limiting reagent of the given reaction.
A limiting reagent is the one which dictates when the reaction will be over. In other words this is the reagent which gets used up first. As the given chemical reaction only has two reactants, if we find the limiting reagent then the other one will be our required answer.
To do this we have to find out the number of moles of reactants provided in the question. The formula for the same is as below:
\[No.\text{ }of\text{ moles}=\dfrac{Given\text{ Mass}}{Molar\text{ Mass}}\]
Using the above formula:
- No. of moles of magnesium $=\dfrac{1g}{24g}=0.0416$ moles
- No. of moles of oxygen $=\dfrac{0.56}{32}=0.0175$ moles
From the reaction equation mentioned above it can be derived that for every mole of oxygen two moles of magnesium is used. We have calculated that $0.0175$ moles of oxygen are given. So the required number of moles of magnesium is-
$0.0175\times 2=0.035$ moles
But we have $0.0416$ moles of magnesium. This means magnesium is in excess.
The excess moles of magnesium is-
$0.0416-0.035=0.0066$ moles
Therefore the amount of magnesium that will be left after the completion of reaction is:
\[0.0066\times 24=0.1584g\]
So the amount of magnesium left after the reaction completion is $0.1584g$
Note: The equation of the reaction must be balanced because all the calculations depend on it.
You could have also done the question by calculating the amount of reactants that will be required in grams, as the final answer is in the same unit.
A mole is itself a unit which is represented as “mol” or “mols” and therefore should not be omitted.
Complete step-by-step answer:
Let us first write the complete reaction with the given reactants and conditions in mind. It is as follows:
\[2Mg(s)+{{O}_{2}}(g)\to 2MgO(s)\]
Here two moles of magnesium reacts with one mole oxygen to form two moles of magnesium oxide. We have to find out which reactant will be left in excess after the reaction is complete. For this we have to first find the limiting reagent of the given reaction.
A limiting reagent is the one which dictates when the reaction will be over. In other words this is the reagent which gets used up first. As the given chemical reaction only has two reactants, if we find the limiting reagent then the other one will be our required answer.
To do this we have to find out the number of moles of reactants provided in the question. The formula for the same is as below:
\[No.\text{ }of\text{ moles}=\dfrac{Given\text{ Mass}}{Molar\text{ Mass}}\]
Using the above formula:
- No. of moles of magnesium $=\dfrac{1g}{24g}=0.0416$ moles
- No. of moles of oxygen $=\dfrac{0.56}{32}=0.0175$ moles
From the reaction equation mentioned above it can be derived that for every mole of oxygen two moles of magnesium is used. We have calculated that $0.0175$ moles of oxygen are given. So the required number of moles of magnesium is-
$0.0175\times 2=0.035$ moles
But we have $0.0416$ moles of magnesium. This means magnesium is in excess.
The excess moles of magnesium is-
$0.0416-0.035=0.0066$ moles
Therefore the amount of magnesium that will be left after the completion of reaction is:
\[0.0066\times 24=0.1584g\]
So the amount of magnesium left after the reaction completion is $0.1584g$
Note: The equation of the reaction must be balanced because all the calculations depend on it.
You could have also done the question by calculating the amount of reactants that will be required in grams, as the final answer is in the same unit.
A mole is itself a unit which is represented as “mol” or “mols” and therefore should not be omitted.
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