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Ammonia

Match the amo...

Question

Match the amount of reactant given in column I with neutralisation reactions given in column II.

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Solution

(A) $1$ . mEq of NaOH $= 200 \times 0.5 = 100 m E q \equiv 100 m E q$ of $H_{2} S O_{4}$
2. $4.9 g H_{2} {+ 6 S O}_{4} = \frac{4.9}{98} \times 10^{3}$
$= 50 m m o l \equiv 100 m E q$ (n factor $= 2)$
$\equiv 200 m m o l$ of O atoms. $(1 m o l H_{2} S O_{4} \equiv 4 m o l O a t o m s)$
3.. Highest oxidation state of $S = + 6$
4. $H y b r i d i s a t i o n = s p^{3}$ ; geometry and shape $\Rightarrow T e t r a h e d r a l$
(B) $4.9 g H_{3} P O_{4} \equiv \frac{4.9}{98} \times 1000 = 50 m m o l$
$= 50 \times 3$ (n factor) $\equiv 150 m E q$
$2. 50 m m o l$ of $H_{3} P O_{4} \equiv 50 \times 4 = 200 m m o l$ of O atoms).
( $1 m o l H_{3} P O_{4} \equiv 4 m o l$ O atoms)
$1$ . But $150$ mEq $H_{3} P O_{4} \neq m E q$ of NaOH
$5$ . Hybridisation is $s p^{3}$ , so geometry and shape is tetrahedral.
$3$ . Maximum oxidation state of $P = + 5$
(C) $1. 4.9 g H_{2} C_{2} O_{4} \equiv \frac{4.9}{90} \times 1000$
$= 50 m m o l = 100 m E q$ (n factor $= 2$ )
2. $50 m m o l e s$ of $H_{2} C_{2} O_{4} \equiv 50 \times 4 = 200 m m o l e s$ O atoms
( $1 m o l H_{2} C_{2} O_{4} = 4 m o l$ o atoms)
$\equiv 100 m E q N a O H$
3. Highest oxidation state of $C = + 4$
4. Reacts with oxidising agents such as $M N O_{4}^{⊖}$ and $C r_{2} O_{7}^{2 -}$ .
(D) $1. 5.3 g$ of $N a_{2} C O)_{3} = \frac{5.3}{106} \times 1000$
$= 50 m m o l e s$
2 $N a_{2} C O_{3}$ does not react with NaOH.
3. Highest oxidation states of $C = + 4$
5. $C O_{3}^{2 -}$ , hybridisation is $s p^{2}$ , so geometry and shape is planar.
7. $1 m o l N a_{2} C O_{3} \equiv 3 m o l$ O atoms.

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