Tag: metal oxides and non metal oxides

Questions Related to metal oxides and non metal oxides

$CuO+(X) \rightarrow CuSO _4+H _2O$ 

Here, $(X)$ is:

  1. $CuSO _4$

  2. $HCl$

  3. $H _2SO _4$

  4. $HNO _3$


Correct Option: C
Explanation:

Acid and base react together, neutralize each other and form salt and water.

$Acid + Base \rightarrow Salt + water$
Here $CuO$ acts as a base and the salt formed is $CuSO _4$, along with formation of water $(H _2O)$.
Since the salt formed is $CuSO _4$, the reacting acid should be sulphate ion ($SO _4^{2-}$) donor, and water would not have been formed if $CuO\  and\  CuSO _4$  would have reacted together. Thus $H _2SO _4$ is the acid that will react with $CuO$ to give $CuSO _4\ and\ H _2O$. The reaction occurs as follows :-
$CuO+H _2SO _4\rightarrow CuSO _4+H _2O$
Hence, the correct answer is option (C).

The equation for the action of a dilute acid on a metal oxide is following:

     Metal oxide + dilute acid $\rightarrow$ Metal salt + Water

  1. True

  2. False


Correct Option: A
Explanation:

Given statement is true. Since water is also formed along with salt .

An example is given below.

$CaO+{ 2HNO } _{ 3 }\rightarrow Ca\left( { NO } _{ 3 } \right) _{ 2 }+{ H } _{ 2 }O$

Chlorine is evolved by the action of:

  1. ${ H } _{ 2 }{ SO } _{ 4 } \ on\ NaCl$

  2. $HCl\ on\ { K } _{ 2 }{ SO } _{ 4 }$

  3. $HCl \ on\ KMN{ O } _{ 4 }$

  4. $HN{ O } _{ 3 }\ on\ KCl$


Correct Option: C
Explanation:

Option $(C)$ is the correct answer.
$Chlorine$ is evolved by the action of $\displaystyle HCl$ on $\displaystyle KMnO _4$
$\displaystyle 2KMnO _4+16HCl \rightarrow 2KCl + 2MnCl _2+8H _2O + 5Cl _2$

Identify acidic salt which can react with one mole of base among the following.

  1. $(NH _{4}) _{2}CO _{3}$

  2. $Na _{2}HPO _{3}$

  3. $NaH _{2}PO _{3}$

  4. $CH _{3}COOK$


Correct Option: C
Explanation:

$H _3PO _3$ is a dibasic acid. In its structure, one oxygen atom is bonded to $P-$ atom via a double bond. $2$ $O-$atoms are bonded to $H-$atoms causing acidity. The third $H$ is directly attached to $P-$ atom responsible for its reducing behavior. 

In $NaH _2PO _3$, one acidic $H$ is replaced by $Na$. Only one $H$ is left which causes the acidic behavior in the salt and hence, will react with one mole of base.