Question

20kg of N2 and 3kg of H2 are mixed to produce NH3.The amount of NH3 formed is?

Answer 1

Answer:

See below.

Explanation:

20 kg N2 and 3 kg H2 is mixed to produce NH3.

The balanced equation is,

$N_2 + 3H_2 \to 2NH_3$

Now, we have 20 kg of N2 which in moles is,

$= \frac{2 \times 10^4}{28} = 7.14 \times 100 = 714 \ mol$

Similarly for 3 kg H2, we have,

$= \frac{3000}{2} = 1500 \ mol$

According to the balanced reaction, we need 1 mole of N2 to react with 3 moles of H2, so for 714 moles of N2, we require,

$714 \times 3 = 2142 \ mol$

which is greater than the amount of H2 we have. Hence H2 is the limiting reagent.

So, 3 moles of H2 produce 2 moles of NH3. 1500 moles will produce,

$= \frac{1500 \times 2}{3} = 1000 \ mol$

moles of NH3.

The mass of NH3 produced,

$1000 \times 17 = 17000 = 17 \ kg$

Therefore, 17 kg NH3 will be produced.

Answer 2

Explanation:

consider hydrogen= 2 and nitrogen = 14 ,And here are 3 hydrogen so 2 * 3=6, 6:14 = 3:7 if 3 kg is given and our ratio is too 3 kg so 7 kg of n2 will make compound and compound mass would be 10 kg

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