Tag: nitrogen and sulfur

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Multiple choice chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

Consider the equilibrium $NH _{4}Cl (s)\rightleftharpoons NH _{3}(g) + HCl(g)$. An inert gas is added to the system at constant volume and temperature. The correct statement(s) is/ are:

  1. the partial pressures of $NH _{3}$ and $HCl$ in the system will increase
  2. the partial pressures of $NH _{3}$ and $HCl$ in the system will remain the same
  3. the partial pressures of $NH _{3}$ and $HCl$ in the system will decrease

  4. the entropy of the system will increase
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

$NH _4$ $+$ $Cl(s)$ $\rightleftharpoons$ $NH _3 (g)$ $+$ $HCl$.

Noble gas are  inert hence adding inert gas doesn't effect the partial pressure.
This is because noble gas gets added to both reactant and product.
$X (g)$ $+$ $NH _4Cl (S)$ $ \rightleftharpoons $ $NH _3(g)$ $+$ $HCl(g)$ $+$ $X(g)$
Where X=Noble gas
Hence the partical presence of $NH _3$ and $HCl$ will remain same.
The amount of gas in the system will increase after adding inert gas.
Since gases has high entropy than liquid and solid, the entropy  of the system will increase.  

Multiple choice chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

Magnesium on heating to redness in an atmosphere of $N _{2}$ and then on treating with $H _{2}O$ gives?

  1. $NH _{3}$
  2. $H _{2}$
  3. $N _{2}$
  4. $O _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Megnesium when heated with $N _2$ in atmosphere it forms magnesium nitride($Mg _3N _2$) which on hydrolysis(reaction with water) gives ammonia.
$3Mg+N _2\rightarrow Mg _3N _2$
$Mg _3N _2+6H _2O\rightarrow 3Mg(OH) _2+2NH _3$.
Hence option A is correct.
Multiple choice chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

In Haber's process $50.0\ g$ of $N _{2}(g)$ and $10.0\ g$ of $H _{2}(g)$ are mixed to produced $NH _{3}(g)$. What are the number of moles of $NH _{3}(g)$ formed?

  1. $3.33$
  2. $2.36$
  3. $2.01$
  4. $5.36$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$N _2 + 3H _2 \rightarrow2NH3$
Here.
28 g $N _2$ reacts with 6 g of $H _2$
So,
50 g of $N _2$ react with $\dfrac{6}{28}\times50 g -H _2 = 10.7 g$ but we have only 10g so $H _2$ is limiting reagent..

6 g $H _2$ gives 34g $NH _3$
10g $H _2$ will give $\dfrac{34}{6}\times10=56.67g -NH _3$
So number of moles of $NH _3$ formed is,
$=\dfrac{Given-mass}{Molar-mass}=\dfrac{56.67}{17}=3.33-moles$