Question

Which of the following statements is/are correct?
i) $4Li + {O_2} \to 2L{i_2}O$
$21.0g$ of lithium reacts with $32.0g$ of ${O_2}$.
(ii) $2K + C{l_2} \to 2KCl$
$3.9g$ of $K$ reacts with $4.26g$ of $C{l_2}$.
[the atomic weight of $Li = 7g$ and $K = 39g$ while molecular weight of $Li{O_2}$ $= 30gmo{l^{ - 1}}$ and $KCl = 74.5gmo{l^{ - 1}}$ ]
A. ${O_2}$ is in excess in reaction (i).
B. $45.0g$ of $Li{O_2}$ is formed in reaction (i).
C. $C{l_2}$ is in excess in reaction (ii).
D. $7.45g$ of $KCl$ is formed in reaction (ii).

Answer 1

In the given question we are asked to find the limiting reagent so for that we can first find out the moles of the given elements and using these moles we can find the limiting reagents.

Complete answer:
As we know that one mole is defined as the ratio of the given mass of the element to the molecular mass of that element. So in the first option when lithium reacts with oxygen it results in the formation of lithium oxide, and we are given that $21.0g$ of lithium reacts with $32.0g$ of ${O_2}$ thus we can find the moles of lithium using:
$moles = \dfrac{{given\;mass}}{{molecular\;mass}}$
$\Rightarrow moles\;of\,Li = \dfrac{{21}}{7} \\\ \Rightarrow moles = 3.0mol \\\$
Similarly we can find out the moles of oxygen:
$\Rightarrow moles\;of\,{O_2} = \dfrac{{32}}{{32}} \\\ \Rightarrow moles = 1.0mol \\\$
Now, $4Li + {O_2} \to 2L{i_2}O$ shows that $4$ moles of Lithium are required to react with $1$ mole of oxygen to give $2$ mole of lithium oxide. And we just found out that $3.0$ moles of lithium are given and these will react with $\dfrac{3}{4}$ moles of oxygen only, so we can say that $1 - \dfrac{3}{4} = 0.25$ moles of oxygen is in excess and lithium is limiting reagent.
So the first statement is correct.
Weight of $Li{O_2}$ formed will be: $= \dfrac{{3 \times 1 \times 30}}{2} = 45.0g$.
Therefore, the second statement is also correct.
In the second option, we are given that potassium reacts with chlorine to given potassium chloride and we are given that $3.9g$ of $K$ reacts with $4.26g$ of $C{l_2}$, so we can calculate the moles of potassium using:
$moles = \dfrac{{given\;mass}}{{molecular\;mass}}$
$\Rightarrow moles\;of\,K = \dfrac{{3.9}}{{39}} \\\ \Rightarrow moles = 0.1mol \\\$
Similarly we can find out the moles of chlorine:
$\Rightarrow moles\;of\,C{l_2} = \dfrac{{4.26}}{{71}} \\\ \Rightarrow moles = 0.06mol \\\$
Now, $2K + C{l_2} \to 2KCl$ shows that $2$ moles of potassium are required to react with $1$ mole of chlorine to give $2$ mole of potassium chloride. And we just found out that $0.1$ moles of potassium are given and these will react with $\dfrac{{0.1}}{2}$ moles of chlorine only, so we can say that $0.1 - \dfrac{{0.1}}{2} = 0.01$ moles of chlorine is in excess and potassium is hence the limiting reagent.
So the third statement is correct.
And, Weight of $KCl$ formed will be: $= 0.1 \times 1 \times 74.5 = 7.45g$.
This is also a true statement.

**Hence, (i), (ii), (iii) and (iv) are correct.

Note:**
Those reagent which is present in lesser amounts as per required from the equation is called limiting reagent as this will be consumed first and it can also decide the amount of product formed as well as amount of other reactants consumed.