Question

A mixture of N2 and H2 is caused to react in a closed container to form Nh3.the reaction ceases before any of the reactant has been totally consumed. At this stage, 2 moles each of N2 and H2 present originally were respectively

Answer 1

N2 + 3H2 → 2NH3

1mol N2 reacts with 3mol H2 to produce 2mol NH3

But you have produced 1.4mol NH3

Rewrite the word equation to reflect this:

0.7Mol N2 reacts with 2.1mol H2 to produce 1.4mol NH3

Original amounts:

N2 = 1.4+0.7 = 2.1 mol N2 originally

H2 = 1.4 + 2.1 = 3.5 mol H2 originally

Answer 2

The gases nitrogen and hydrogen reacts to form ammonia gas.

The balanced chemical reaction is as follows. ${ N }_{ 2 }(g)\quad +\quad 3{ H }_{ 2 }(g)\quad \quad \rightarrow \quad 2{ NH }_{ 3 }(g)\quad$

The reaction stops before the total consumption of either reactant, where the moles of gase present are,

$2.0\quad mole\quad of\quad { NH }_{ 3 },\quad \quad 2.0\quad mole\quad of\quad { N }_{ 2 }\quad and\quad 2.0\quad mole\quad of\quad { H }_{ 2 }.$

$Therefore,\quad 2\quad moles\quad of\quad { NH }_{ 3 }\quad is\quad formed\quad form\quad \quad { N }_{ 2 }\quad and\quad \quad { H }_{ 2 }.$

From the balanced reaction 1 mole of ${ N }_{ 2 }$ and 3 moles of ${ H }_{ 2 }$ produces 2 moles of ${ NH }_{ 3 }$ (1:3:2)

Then to produce 2.0 moles of ${ NH }_{ 3 }$

We require,

$1.5\quad moles\quad of\quad { N }_{ 2 }\quad and\quad 4.5\quad moles\quad of\quad { H }_{ 2 }$

So, the number of moles of ${ N }_{ 2 }$ and ${ H }_{ 2 }$ originally present in the contaier is moles required for forming $N{ H }_{ 3 }$ and moles left after end of the reacion.

$(1.5\quad mol\quad { N }_{ 2 }\quad +\quad 2.0\quad mol\quad { N }_{ 2 }\quad )\quad =\quad 3.5\quad mol\quad { N }_{ 2 }\quad$

$(4.5\quad mol\quad { H }_{ 2 }\quad +\quad 2.0\quad mol\quad { H }_{ 2 })\quad =\quad 6.5\quad \quad mol\quad { H }_{ 2 }$