Question

in reaction n2+3h2 give 2NH3 100g N2 reacts with 20g h2 then find (i) limiting reagent (ii) find amount of nh3 obtained​

Answer 1

Answer:

(i) Limiting reagent is H2 and (ii) amount of NH3 is 6.667mol

Explanation:

N2 + 3H2= 2NH3

So, 1mol of N2 is reacting with 3mol of H2 resulting 2 mol of NH3.

Here, according to the question, we have taken 100g of N2. Molecular weight of N2 is 14*2=28g/mol

So, number of moles of N2 present here is  100g/(28g/mol)= 3.57mol

and the number of moles of H2 is 20/2=10mol as we have taken 20g of h2 ( molecular weight 2g/mol)

10mol of H2 will react only with 10/3= 3.333mol of N2 as they react in 1:3 proportion.

So, (3.57-3.33)=.24mol of N2 will be unreacted.

In this reaction, the amount of H2 is deciding how much N2 should react with it and how much NH3 will be formed. So, H2 is the limiting reagent here.

From the balanced equation we see, 3 mol of H2 form 2mol of NH3

so, the same way, 1 mol H2 forms 2/3 mol of NH3

10mol of H2 will form (10*2)/3= 6.667 mol

Answer 2

Answer:

• Molar mass of N₂ = 28 g/mol
• Molar mass of H₂ = 2 g/mol
• Stoichiometric coefficient of N₂ = 1
• Stoichiometric coefficient of H₂ = 3
• Given mass of N₂ = 100 g
• Given mass of H₂ = 20g

(I)

First let's find the number of moles of given reactants in the reaction:

$\footnotesize \implies \rm n_{ \tiny{N_2}} = \dfrac{100}{28} = \red{ 3.6 \: mol}$

$\footnotesize \implies \rm n_{ \tiny{H_2}} = \dfrac{20}{2} = \red{ 10 \: mol}$

Now, we will find the limiting reagent:

$\tiny \implies \underline{ \boxed{ \purple{\bf Limiting\: reagent= \dfrac{no. \: of \: moles \: of \: reactant \: given \: in \: reaction}{ Stoichiometric \: coefficient \: of \: that \: reactant}}}}$

$\footnotesize \implies \rm L.R_{ \tiny{N_2}} = \dfrac{3.6}{1} = \gray{ 3.6}$

$\footnotesize \implies \rm L.R_{ \tiny{H_2}} = \dfrac{10}{3} = \gray{ 3.4}$

Therefore, Limiting reagent in the given reaction is H₂.

(ii)

$\footnotesize \sf 1N_2 + 3H_2 \longrightarrow 2NH_3$

If 3 moles of H₂ will form 2 moles of NH₃.

Then, 10 moles of H₂ will form moles of NH₃ = $\rm \dfrac{2}{3} \times 10 = \green{6.7 \:moles}$

Now,

1 mole of NH₃ has mass = molar mass of NH₃ = 17 g

Then, 6.7 moles of NH₃ has mass = 17 × 6.7 = 113.9 g

Hence,the amount of NH₃ is obtained = 113.9 g.