In the following reactions, the total number of oxygen atoms in X… — Solution | JEE Chemistry

NVAEasyJEE 2023Stoichiometry & Calculations

In the following reactions, the total number of oxygen atoms in $X$ and $Y$ is: $Na_2O + H_2O \rightarrow 2X$ $Cl_2O_7 + H_2O \rightarrow 2Y$

Answer

Correct answer:5

Standard Method

Given: $Na_2O + H_2O \rightarrow 2X$ $Cl_2O_7 + H_2O \rightarrow 2Y$

Find: The total number of oxygen atoms in $X$ and $Y$ .

The reactions of sodium oxide and dichlorine heptoxide with water are as follows:

Na_2O + H_2O \rightarrow 2NaOH

Cl_2O_7 + H_2O \rightarrow 2HClO_4

So, $X = NaOH$ and $Y = HClO_4$ .

Number of oxygen atoms in $X$ = $1$

Number of oxygen atoms in $Y$ = $4$

Total number of oxygen atoms in $X$ and $Y$ = $1 + 4 = 5$ .

Therefore, the required answer is $5$ .

Taking $X$ and $Y$ as the full products $2NaOH$ and $2HClO_4$ instead of the species $NaOH$ and $HClO_4$ . This is wrong because the question asks for oxygen atoms in $X$ and $Y$ , not in $2X$ and $2Y$ . First identify the compound represented by each symbol, then count oxygen atoms in one formula unit.
Not identifying that oxides react with water to form the corresponding hydroxide or oxyacid. This leads to wrong products and wrong oxygen count. Balance each reaction correctly before assigning $X$ and $Y$ .

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