JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 10)
For a reaction,
N2(g) + 3H2(g) $$ \to $$ 2NH3(g) ;
identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures.
N2(g) + 3H2(g) $$ \to $$ 2NH3(g) ;
identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures.
56g of N2 + 10g of H2
14g of N2 + 4g of H2
28g of N2 + 6g of H2
35g of N2 + 8g of H2
Explanation
Here you have to check every options to find answer.
N2(g) + 3H2(g) $$ \to $$ 2NH3(g)
For dihydrogen (H2) to become a limiting reagent,
$${{moles\,of\,{H_2}} \over {stoichiometric\,coefficient\,of\,{H_2}}}$$ should be
less than $${{moles\,of\,{N_2}} \over {stoichiometric\,coefficient\,of\,{N_2}}}$$.
Here in 56g of N2 + 10g of H2,
Moles of N2 = $${{56} \over {28}}$$ = 2
and Moles of H2 = $${{10} \over 2}$$ = 5
Now by dividing stoichiometric coefficient we get,
For N2 = $${2 \over 1}$$ = 2
For H2 = $${5 \over 3}$$ = 1.67
$$ \therefore $$ Here H2 is the limiting reagent.
N2(g) + 3H2(g) $$ \to $$ 2NH3(g)
For dihydrogen (H2) to become a limiting reagent,
$${{moles\,of\,{H_2}} \over {stoichiometric\,coefficient\,of\,{H_2}}}$$ should be
less than $${{moles\,of\,{N_2}} \over {stoichiometric\,coefficient\,of\,{N_2}}}$$.
Here in 56g of N2 + 10g of H2,
Moles of N2 = $${{56} \over {28}}$$ = 2
and Moles of H2 = $${{10} \over 2}$$ = 5
Now by dividing stoichiometric coefficient we get,
For N2 = $${2 \over 1}$$ = 2
For H2 = $${5 \over 3}$$ = 1.67
$$ \therefore $$ Here H2 is the limiting reagent.
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