If $ {\text{1}}{\text{.08 g}} $ of aluminium and $ {\text{0}}{\text{.99 g}} $ of oxygen are available for this reaction, who is the limiting reactant?
$ {\text{4 Al + 3}}{{\text{O}}_{\text{2}}} \to 2{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} $
A.Al
B. $ {{\text{O}}_{\text{2}}} $
C. $ {\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} $
D.Mole ratios of reactants are perfectly balanced
Answer
607.5k+ views
Hint: The limiting reactant in a chemical reaction is the reactant that is totally consumed when the chemical reaction is completed. The amount of product is limited by this reagent since the reaction cannot continue without it.
Complete step by step answer:
As per the reaction given in the question, 4 moles of aluminium react with 3 moles of oxygen to form 2 mole of aluminium oxide. The atomic weight of aluminium is 27 while the molecular weight of oxygen is 32 and the molecular weight of aluminium oxide is $ \left( {27 \times 2} \right) + \left( {16 \times 3} \right) = 56 + 48 = 104 $ . The amount of aluminium taken for the reaction is $ {\text{1}}{\text{.08 g}} $ while that for oxygen is $ {\text{0}}{\text{.99 g}} $ .
Hence the number of moles of aluminium taken for the reaction is $ \dfrac{{{\text{1}}{\text{.08}}}}{{27}}{\text{ = 0}}{\text{.04}} $ moles,
while the number of moles of oxygen taken for the reaction is equal to
$ \dfrac{{0.99}}{{32}} = 0.031 $ Moles.
As per the stoichiometry of the reaction, for $ 0.04 $ moles of aluminium, $ 0.03 $ moles of oxygen is required. So almost $ 0.001 $ moles of oxygen is left behind.
Hence, the limiting reagent here is aluminium.
So, the correct answer is option A.
Note:
If one or more than one than one reagent is present in excess of the quantities required to react with the limiting reagent then they are called the excess reagents or excess reactants. Hence the concentration of the limiting reagent must be identified to calculate the percentage yield of a reaction since the theoretical yield is defined as the amount of the product obtained when the limiting reagent reacts completely.
Complete step by step answer:
As per the reaction given in the question, 4 moles of aluminium react with 3 moles of oxygen to form 2 mole of aluminium oxide. The atomic weight of aluminium is 27 while the molecular weight of oxygen is 32 and the molecular weight of aluminium oxide is $ \left( {27 \times 2} \right) + \left( {16 \times 3} \right) = 56 + 48 = 104 $ . The amount of aluminium taken for the reaction is $ {\text{1}}{\text{.08 g}} $ while that for oxygen is $ {\text{0}}{\text{.99 g}} $ .
Hence the number of moles of aluminium taken for the reaction is $ \dfrac{{{\text{1}}{\text{.08}}}}{{27}}{\text{ = 0}}{\text{.04}} $ moles,
while the number of moles of oxygen taken for the reaction is equal to
$ \dfrac{{0.99}}{{32}} = 0.031 $ Moles.
As per the stoichiometry of the reaction, for $ 0.04 $ moles of aluminium, $ 0.03 $ moles of oxygen is required. So almost $ 0.001 $ moles of oxygen is left behind.
Hence, the limiting reagent here is aluminium.
So, the correct answer is option A.
Note:
If one or more than one than one reagent is present in excess of the quantities required to react with the limiting reagent then they are called the excess reagents or excess reactants. Hence the concentration of the limiting reagent must be identified to calculate the percentage yield of a reaction since the theoretical yield is defined as the amount of the product obtained when the limiting reagent reacts completely.
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