Question

calculate the mass of the ammonia obtained when 10 grams of N2 reacts with 5g of H2​

Answer 1

Answer:

N2(g) + 3H2(g) —> 2NH3(g)

28g 6g 34g

28 g of N2 reacts with 6g of H2 .

1g of N2 reacts with 6/28 g of H2

2000g of N2 will react with 2000× 6/28

= 428.57 of H2

but H2 is given 1 × 10³g is greater than 428.57 g

N2 is a limiting reagent and limits the production of ammonia

28g of N2 produces 34 g of NH3

1g of N2 produces 34/28 g of NH3

2000g of N2 will produce 34/28 × 2000

= 2428.57 g of NH3 .

(ii) H2 is excess so it remains unreacted

(iii) Mass of unreacted dihydrogen = 1000-428.57

=571.43g

Explanation:

plz mark as brainlist...

Answer 2

Answer:

N2+3H2→2NH3

28 grams of N2 reacts with 6 grams of H2

10 grams of N2 reacts with 'x' grams of H2

x=10*6/28≈2.142 grams

10 grams of N2 can only react with 2.142 grams of H2 so, remaining is excess reagent

N2 is limiting reagent

now,

wkt

28 g of N2 gives 34 g of NH3

10 g of N2 gives 'x' g of NH3

x=340/28=12.142 g of NH3