Which of the following statements is/are correct?i. 21.0 of lithium reacts with 32.0 g of O24Li+O2⟶2Li2O ii. 3.9 g of K reacts with 4.26 g of Cl22K+Cl2⟶2KCl[Atomic weight of Li = 7 and K = 39. Mw of Li2O = 30 and KCI =74.5 g mol-l)

Moles of Li = 217=3.0 molMoles of O2 =3232=3.0 mol1. Since (3.0 mol Li) 1 mol O24 mol Li=34=0.75 mol of O2 is required, therefore (1.0-0.75) =0.25 mol of O2 is in excess. Hence, Li is the limiting reagent 2. Weight of Li2O formed =(3.0 mol Li)1 mol Li2O2 mol Li30 g Li2Omol Li2O=3×1×302=45.0 g Li2O 3. Moles of K = 3.939=0.1 mol Moles of Cl2=4.2671=0.06 mol Since 0.12=0.05 mol of Cl2 is required. Therefore, (0.06 - 0.05) =0.01 mol of Cl2 is in excess Hence K is the limiting reagent. 4. Weight of KCl formed =(0.1 mol K)1 mol KClmol K74.5 g KClmol KCl=0.1×1×74.5=7.45 g of KCl

Which of the following statements is/are correct?i. 21.0 of lithium reacts with 32.0 g of O24Li+O2⟶2Li2O ii. 3.9 g of K reacts with 4.26 g of Cl22K+Cl2⟶2KCl[Atomic weight of Li = 7 and K = 39. Mw of Li2O = 30 and KCI =74.5 g mol-l)

Moles of Li = 217=3.0 molMoles of O2 =3232=3.0 mol1. Since (3.0 mol Li) 1 mol O24 mol Li=34=0.75 mol of O2 is required, therefore (1.0-0.75) =0.25 mol of O2 is in excess. Hence, Li is the limiting reagent 2. Weight of Li2O formed =(3.0 mol Li)1 mol Li2O2 mol Li30 g Li2Omol Li2O=3×1×302=45.0 g Li2O 3. Moles of K = 3.939=0.1 mol Moles of Cl2=4.2671=0.06 mol Since 0.12=0.05 mol of Cl2 is required. Therefore, (0.06 - 0.05) =0.01 mol of Cl2 is in excess Hence K is the limiting reagent. 4. Weight of KCl formed =(0.1 mol K)1 mol KClmol K74.5 g KClmol KCl=0.1×1×74.5=7.45 g of KCl

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Which of the following statements is/are correct?
i. 21.0 of lithium reacts with 32.0 g of O₂
$4 Li + O_{2} ⟶ 2 {Li}_{2} O$
ii. 3.9 g of K reacts with 4.26 g of Cl₂
$2 K + {Cl}_{2} ⟶ 2 KCl$
[Atomic weight of Li = 7 and K = 39. Mw of Li₂O = 30 and KCI =74.5 g mol^-l)

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45.0 g of Li₂O is formed is reaction (i).

7.45 g of KCl is formed is reaction (ii).

In reaction (i), O₂ is in excess

In reaction (ii), Cl₂ is in excess

answer is A, B, C, D.

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Detailed Solution

Moles of Li = $\frac{21}{7} = 3.0 mol$

Moles of O₂ = $\frac{32}{32} = 3.0 mol$

$1 . Since (3.0 mol Li) (\frac{1 mol O_{2}}{4 mol Li}) = \frac{3}{4} = 0.75 mol of O_{2} is required, therefore (1.0 - 0.75) = 0.25 mol of O_{2} is in excess . Hence, Li is the limiting reagent 2 . Weight of {Li}_{2} O formed \begin{array}{l} = (3.0 mol Li) (\frac{1 mol {Li}_{2} O}{2 mol Li}) (\frac{30 g {Li}_{2} O}{mol {Li}_{2} O}) \\ = \frac{3 \times 1 \times 30}{2} = 45.0 g {Li}_{2} O \end{array} 3 . Moles of K = \frac{3.9}{39} = 0.1 mol Moles of {Cl}_{2} = \frac{4.26}{71} = 0.06 mol Since (\frac{0.1}{2}) = 0.05 mol of {Cl}_{2} is required . Therefore, (0.06 - 0.05) = 0.01 mol of {Cl}_{2} is in excess Hence K is the limiting reagent . 4 . Weight of KCl formed \begin{array}{l} = (0.1 mol K) (\frac{1 mol KCl}{mol K}) (\frac{74.5 g KCl}{mol KCl}) \\ = 0.1 \times 1 \times 74.5 = 7.45 g of KCl \end{array}$