Question

What is the maximum amount of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} $ which could be formed from the reaction of $ 12.71\text{ g }Al $ and $ 10.09\text{ g }CuS{{O}_{4}} $ ?
 $ Al+CuS{{O}_{4}}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+Cu $ .

Answer 1

Hint: Limiting reagent: It is the reagent or reactant present in the chemical reaction, which is completely consumed after the formation of the product and hence, is responsible to decide that when a chemical reaction will stop and plays a major role to determine the amount of each molecule formed as a product.

Complete answer:
The chemical reaction given in the question is as follows:
 $ Al+CuS{{O}_{4}}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+Cu $
Step-1: Balancing the given chemical reaction:
On comparing the number of atoms of each element in reactants as well as products and adding the required stoichiometric coefficients, the balance reaction is as follows:
 $ 2Al+3CuS{{O}_{4}}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3Cu $
Step-2: Finding the limiting reagent in the given chemical reaction:
Given mass of $ CuS{{O}_{4}}=10.09g $
Molar mass of $ CuS{{O}_{4}}=160\ gmo{{l}^{-1}} $
 $ \because $ number of moles $ =\dfrac{\text{given mass}}{\text{molar mass}} $
 $ \therefore $ number of moles of $ CuS{{O}_{4}}=\dfrac{10.09}{160}\Rightarrow 0.063\ moles $
Given mass of $ Al=12.71g $
Molar mass of $ Al=27\ gmo{{l}^{-1}} $
 $ \therefore $ number of moles of $ Al=\dfrac{12.71}{27}\Rightarrow 0.47\,moles $
As per reaction,
 $ 2 $ moles of $ Al $ reacts with $ \Rightarrow 3 $ moles of $ CuS{{O}_{4}} $
Therefore, $ 0.47 $ moles of $ Al $ will reacts with $ \Rightarrow \dfrac{3}{2}\times 0.47=0.705 $ moles of $ CuS{{O}_{4}} $
But, the number of moles of $ CuS{{O}_{4}} $ present in the reaction $ =0.063\ moles $
Hence, $ CuS{{O}_{4}} $ is the limiting reagent for the given chemical reaction.
Step-3: Calculation of number of moles of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} $ formed after reaction:
As per given reaction,
 $ 3 $ moles of $ CuS{{O}_{4}} $ reacts to form $ \Rightarrow 1 $ mole of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} $
Therefore, $ 0.063 $ moles of $ CuS{{O}_{4}} $ will reacts to form $ \Rightarrow \dfrac{1}{3}\times 0.063=0.021 $ moles of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} $
Hence, number of moles of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} $ formed $ =0.021 $ moles
Step-4: Calculation of mass of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} $ formed after reaction:
Molar mass of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}}=342.15\text{ gmo}{{\text{l}}^{-1}} $
Number of moles of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}}=0.021 $ moles
 $ \because $ number of moles $ =\dfrac{\text{given mass}}{\text{molar mass}} $
Therefore, mass $ = $ number of moles $ \times $ molar mass
Substituting values:
Mass $ =0.021\times 342.15\Rightarrow 7.18\text{ g} $
Hence, mass of $ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} $ formed $ =7.18\ \text{g} $ .

Note:
Ensure that the chemical reaction is balanced i.e., number of atoms of each element are same in reactants as well as product, before finding the limiting reagent and the amount of final product is always evaluated according to the amount of limiting reagent present in the reaction.