Tag: nitrogen and sulfur

Questions Related to nitrogen and sulfur

Ammonia evolved from the treatment of $0.30$g of an organic compound for the estimation of nitrogen was passed in $100$ mL of $0.1$M sulphuric acid. The excess of acid required $20$ mL of $0.5$M sodium hydroxide solution for complete neutralization. The organic compound is:

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. thiourea

  2. benzamide

  3. urea

  4. acetamide

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Correct Option: C
Explanation:

Organic compound+ $H _2SO _4\longrightarrow$

Using Kjeldhal Process to calculate the percentage of nitrogen organic compound
Let unreacted $0.1M(=0.2N)H _2SO _4=Vml$
$\therefore, 20ml$ of $0.5M$ $NaOH=Vml$ of $0.2g$ $NH _2SO _4$
$\therefore 20 \times 0.5=V\times 0.2$
or, $V=50ml$
used $H _2SO _4=100-50=50ml$
$\therefore,$ % nitrogen=$14NV/w$
where $N$=normality of $H _2SO _4$
$V$=Volume of $H _2SO _4$ used
% nitrogen=$1.4\times 0.5\times 50\times 0.3=46.67$ %
Urea=$NH _2CONH _2=28\times 100/60$
$=46.67$ %
Thus, the organic compound is urea.

Ammonium ion is detected by the addition of an alkali.

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. True

  2. False

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Correct Option: A
Explanation:

Hydrogen chloride gas gives dense white fumes (of ammonium chloride) with ammonium hydroxide.
 $\displaystyle HCl + NH _4OH \rightarrow NH _4Cl + H _2O$
Ammonium ion is detected by the addition of an alkali.
Ammonium salts when heated with alkali, give ammonia.
 $\displaystyle NH _4Cl+NaOH \rightarrow NaCl + NH _3 \uparrow +H _2O$
Evolution of ammonia gas can be detected by paper dipped in $\displaystyle CuSO _4$ solution. Ammonia changes the colour of this paper to deep blue due to formation of tetrammine copper complex.
$\displaystyle CuSO _4+4NH _3 \rightarrow [Cu(NH _3) _4]SO _4$

Freshly prepared pure dilute solution of sodium in liquid ammonia :

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. shows copper-bronze colour.

  2. occupy larger volume than that from the sum of the volumes of $Na$ and $NH _{3}(1)$.

  3. reduced the $GeH _{4}$ to $GeH _{2}^{-}$.

  4. produced sodium amide and hydrogen gas with rusty iron wire.

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Correct Option: A,D
Explanation:

In dilute solutions  of $Na$ in ammonia, the solvated electrons have a deep blue colour.

In more concentrated solutions, the solution takes on a bronze/copper colour.

The solution of sodium in liquid ammonia is quite stable on its own.

You need to add a small amount of a catalyst such as ${ FeCl } _{ 3 }$ for the sodium to react with the ammonia to form sodium amide and hydrogen gas.

$2N{ a } _{ (s) }+2N{ H } _{ 3(l) }\underrightarrow { \quad FeC{ l } _{ 3 } } 2NaN{ H } _{ 2 }(N{ H } _{ 3 })+{ H } _{ 2(g) }$
conclusion : hence the option (A) and(D) is correct.

Which of the following are correct statements?

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. $NH _3$ can be used as a refrigerant

  2. NO is heavier than $O _2$

  3. Nitrogen diffuses fastre than $O _2$

  4. $NO _2$ is a colourless gas

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Correct Option: A
Explanation:

Ammonia is amongst the oldest of all there frigerants and still used widely in the refrigeration applications. It is also the only refrigerant outside the halo-carbons group, still being used to a great extent.Ammonia refrigerant is commonly known as R717 and its chemical formula is $NH _3$.

Which of the following can be classified as a molecular hydride ?

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. $LiH$

  2. $\mathrm { NH } _ { 3 }$

  3. $\mathrm { NiHO } _ { 6 }$

  4. $\mathrm { MgH } _ { 2 }$

Show answer
Correct Option: B
Explanation:
Hydride is the product when hydrogen reacts with any other elements, except the noble gases. So hydrides are compounds where one atom is hydrogen to another more electropositive element. A hydride is an anion of hydrogen $(H^-)$.
Bonding between hydrogen & other elements is always covalent. General chemical formula of a hydride is $MH _{x}$ where   $M=$ other element with hydrogen makes a binary compound. $x=$ no. of hydrogen atoms.
Covalent or Molecular hydrides
Hydrogen atoms from a covalent bond with $p-$block atoms. General chemical formula$-XH _{(8-n)}$ where $n=$no. of electrons the element has in its atomic shell.
$NH _{3}$ is $c$ molecular hydride
$3H _{2(g)}+N _{2(g)}\rightarrow 2NH _{3(g)}$
Answer is $B)\ NH _{3}$

When chlorine reacts with a gas X, an explosive inorganic compound Y is formed. Then X and Y will be :

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. $\displaystyle X=O _{2}$ and $\displaystyle Y=NCl _{3}$

  2. $\displaystyle X=NH _{3}$ and $\displaystyle Y=NCl _{3}$

  3. $\displaystyle X=O _{2}$ and $\displaystyle Y=NH _{4}Cl$

  4. $\displaystyle X=NH _{3}$ and $\displaystyle Y=NH _{4}Cl$

Show answer
Correct Option: B
Explanation:

When $NH _3$ reacts with excess of $Cl$ an explosive substance $NCl _3$ is formed.
$NH _3 + 3Cl _2 \rightarrow NCl _3 + 3HCl$
X is $NH _3$ and Y is $NCl _3$. 

Ammonia cannot be dried over regular dehydrating agent like: 

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. $H _{2}SO _{4}$

  2. $P _{2}O _{5}$

  3. Anhydrous $CaCl _{2}$

  4. None of these

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Correct Option: A
Explanation:

Ammonia cannot be dried by passing over conc. sulphuric acid because ammonia gas reacts chemically with conc. sulphuric acid to form ammonium sulphate.
2NH3 + H2SO4 (conc.) $\rightarrow$(NH4)2SO4.

Identify $(E)$ and $(F)$ in the following:
$CaCO _3\rightarrow \underset{(g)}{(A)}+\underset{(solid)}{(B)}$

$(B)+H _2O\rightarrow (C)\overset{(A)}{\rightarrow}(D)\overset{(A)excess}{\rightarrow}(E)$

$(A)+NH _3\overset{\Delta}{\rightarrow}(F)$

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. $(E)\ Ca(HCO _3) _2;\ (F)\ NHCONH _3$

  2. $(E)\ Ca(HCO _3); \ (F)\ NH _3CONH$

  3. $(E) \ Ca(HCO _3) _2; \ (F)\ NH _2CONH _2$

  4. None of these

Show answer
Correct Option: C
Explanation:

$CaCO _3\rightarrow \underset{(g)}{CO _2}+\underset{(solid)}{CaO}$
$CaO+H _2O\rightarrow (Ca(OH) _2)\overset{(CO _2)}{\rightarrow}(CaCO _3)\overset{(CO _2)excess}{\rightarrow}(Ca(HCO _3) _2)$
$(CO _2)+NH _3\overset{\Delta}{\rightarrow}(NH _2CONH _2)$
Hence option C is correct.

$NH _3+ O _2 \longrightarrow$ (B) (brown fumes)

Identify the compound B.

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. $NO$

  2. $NO _2$

  3. $NO _3$

  4. $HNO _2$

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Correct Option: B
Explanation:

Reaction:
$4NH _3+ 7O _2 \longrightarrow 4NO _2 + 6H _2O$ 

$NH _3 + I^- \longrightarrow$ (E) (violet vapours)
What is (E) ?

chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

  1. $I _2$

  2. $I _2^{-}$

  3. $I$

  4. $I _3^{-}$

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Correct Option: A
Explanation:

Reaction:
$NH _3 + I^- \longrightarrow  I _2$ (violet vapours)

Some steamy fumes of $HI$ is formed in this reaction with lots of violet vapours of $I _2$.