XBarryX, nope I don't mind.
I am 12 years old, I like horses and cats, and I love to read warriors.
Answers
Answer:
honey what a sweet name
so honey I to love warrior like in mahabharata
Answer:
Given :-
56 g of Nitrogen reacts with 22 g of hydrogen to form ammonia.
To Find :-
The limiting reagent of the reaction.
The amount of reactant in excess.
The amount of ammonia formed.
Solution :-
\clubsuit♣ Balanced Equation :
\bigstar\: \: \sf\boxed{\bold{N_2 + 3H_2 \longrightarrow 2NH_3}}★N2+3H2⟶2NH3
Now, we have to find the number of moles of nitrogen and hydrogen :
{\normalsize{\bold{\purple{\underline{\leadsto\: In\: case\: of\: Nitrogen\: :-}}}}}⇝IncaseofNitrogen:−
\begin{gathered}\implies \sf \bold{\pink{Number\: of\: moles =\: \dfrac{Mass}{Molar\: Mass}}}\\\end{gathered}⟹Numberofmoles=MolarMassMass
\implies \sf Number\: of\: moles =\: \dfrac{\cancel{56}}{\cancel{28}}⟹Numberofmoles=2856
\implies \sf Number\: of\: moles =\: \dfrac{2}{1}⟹Numberofmoles=12
\implies \sf\bold{\green{Number\: of\: moles =\: 2\: moles}}⟹Numberofmoles=2moles
{\normalsize{\bold{\purple{\underline{\leadsto \: In\: case\: of\: hydrogen\: :-}}}}}⇝Incaseofhydrogen:−
\implies \sf Number \: of\: moles =\: \dfrac{\cancel{22}}{\cancel{2}}⟹Numberofmoles=222
\implies \sf Number\: of\: moles =\: \dfrac{11}{1}⟹Numberofmoles=111
\implies \sf \bold{\green{Number\: of\: moles =\: 11\: moles}}⟹Numberofmoles=11moles
Again,
\clubsuit♣ Balanced Equation :
\begin{gathered}\bigstar\: \: \: \sf\boxed{\bold{28\: g\: of\: N_2 \longrightarrow 34\: g\: of\: NH_3}}\\\end{gathered}★28gofN2⟶34gofNH3
Given that :
\mapsto↦ 56 g of Nitrogen
\implies \sf 56 \times \dfrac{34}{28}⟹56×2834
\implies \sf \dfrac{1904}{28}⟹281904
\implies \sf \dfrac{68}{1}⟹168
\implies \sf\bold{\red{68\: g}}⟹68g
\rule{150}{3}
1) The limiting reagent of the reaction :
➲ The limiting reagent of the reaction is Nitrogen
2) The amount of reactant in excess :
➲ The amount of reactant in excess is Hydrogen
3) The amount of ammonia formed :
➲ The amount of ammonia formed is 68 grams