Hindi, asked by poornimasaliyan1, 3 months ago

Please Answer fast today is my exam ​

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Answered by Anonymous
2

Answer:

कृष्ण- श्याम, कन्हैया, नंदनंदन, कसारि, गोपीवल्लभ.

Explanation:

Hope it's helpful to you✌️

Answered by XBarryX
0

Answer:

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

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