Question

Suppose 316.0g aluminium sulphide reacts with 493.0g of water. What mass of the excess reactant remains? The unbalanced equation is:
\[A{{l}_{2}}{{S}_{3}}+{{H}_{2}}O\to Al{{(OH)}_{3}}+{{H}_{2}}S\]

Answer 1

Hint The answer is based on the balancing of the chemical equation given and then finding the excess reagent and the limiting reagent present and calculating the mass of the excess reagent by subtracting mass of limiting reagent from the excess reagent and multiplying the answer with its respective number of moles.

Complete step – by – step answer:
In the lower classes of chemistry, we have studied the basic concepts of chemistry which tells us about the definitions of basic terms such as the limiting reagent and excess reagent, molar mass calculation and many other concepts.
Let us now look in detail about finding the excess reagent by balancing the given chemical equation and followed by calculation of excess mass of the excess reagent.
- Limiting reagent is the one which gets consumed completely in the given reaction and the excess reagent is the one which reacts with limiting reagent and the remaining amount comes out to be excess.
- In the above given equation, let us first balance the given equation and that will be,
\[A{{l}_{2}}{{S}_{3}}+6{{H}_{2}}O\to 2Al{{(OH)}_{3}}+3{{H}_{2}}S\]
Thus, the above balanced equation shows that one mole of aluminium sulphide reacts with 6 moles of water to give the product.
Here, according to the data, the amount of aluminium sulphide given is 316.0g and that of water is 493.0g.
Now, let us calculate the quantity of the reagents per mole of a substance that will be done by using the formula\[\dfrac{m}{M}\]
Now, calculating this for each of the reactant we have,
\[A{{l}_{2}}{{S}_{3}}=\dfrac{316}{150}=2.10g/mol\] and \[{{H}_{2}}O=\dfrac{493}{18}=27.38g/mol\]
To find which one among these two is the excess reagent, we shall divide the value by its respective number of moles and that is,
\[A{{l}_{2}}{{S}_{3}}=\dfrac{2.10}{1}=2.10g\] and \[{{H}_{2}}O=\dfrac{27.38}{6}=4.56g\]
Therefore, 2.10g of aluminium sulphide needs 4.56g of water to give the required product and by this data, we can say that water is present in excess and acts as an excess reagent and aluminium sulphide is the limiting reagent.
Thus, the excess mass present will be = $18\times \left( 4.56-2.10 \right)=44.37g$

Thus, the correct answer is the excess reactant has the mass of 44.37g

Note: Note that limiting reagent method is most useful when there are only two reactants and this limiting reactant/ reagent can be derived by the comparison of the amounts of products which can be formed from each reactant.