Question

dinitrogen and dihydrogen react with each other to produce ammonia .Calculate the mass of NH3 produced when 2*10^3g nitrogen react with 1*10^3g hydrogen

Answer 1

Answer:

The amount of ammonia formed when $2\ *\ 10^3$ g of dinitrogen reacted with $1\ *\ 10^3$ g of dihydrogen is $2.43\ *\ 10^3$ g of ammonia·

Explanation:

The balanced chemical equation for the given reaction is,

        $N_{2}_(_g_)\ \ +\ \ 3H_{2}_(_g_)$    →    $2NH_{3}_(_g_)$

From the reaction, it is clear that,

 One molecule of dinitrogen reacted with three molecules of dihydrogen to give two molecules of ammonia as a product·

Given that,

The weight of nitrogen reacted  $=\ \ \ 2$ × $10^{3}$ g

The weight of hydrogen reacted $=\ \ 1$ × $10^{3}$ g

We know that,

The molecular mass of dinitrogen   $=\ \ 28$ g

The molecular mass of dihydrogen $=\ \ 2$ g

The molecular mass of ammonia     $=\ \ 17$ g

Three moles of dihydrogen  $=\ \ 3$ × $2$ g  $=\ \ 6$ g

Two moles of ammonia  $=\ \ 2$ × $17$ g  $=\ 34$ g

So,

$28$ g of dinitrogen gives $34$ g of ammonia

Then,

$2\ *\ 10^3$ g of dinitrogen gives $\frac{2\ *\ 10^3\ g\ \ *\ \ 34\ g}{28\ g}$

That is,

          $2.43\ *\ 10^3$ g of ammonia

Here, dinitrogen is the limiting reagent and dihydrogen is in excess·

The amount of dihydrogen that is excess is,

 ⇒    $\frac{2\ *\ 10^3\ *\ 6}{28}$

 ⇒      $428.5$ g

 ∴  The excess amount  $=\ 1\ *\ 10^3\ g \ -\ 428.5\ =\ 571.5$ g of dihydrogen

The reaction goes on in accordance with the limiting reagent· So in order to find out the amount of ammonia formed, we should relate it with the limiting reagent, that is dinitrogen·

Hence,

The amount of ammonia formed when $2\ *\ 10^3$ g of dinitrogen reacted with $1\ *\ 10^3$ g of dihydrogen is $2.43\ *\ 10^3$ g of ammonia·