Question

The reaction for the decomposition of ammonia (NH3) can be written as shown. If a student starts with 21.7 g of ammonia, how many grams of hydrogen gas (H2) will be produced by the reaction?

Answer 1

3.88 grams will be released I believe

Answer 2

Answer:- 3.83 g of hydrogen gas are formed.

Solution:- It is a stoichiometry problem and could easily be solved using dimensional analysis. The balanced equation for the decomposition of ammonia gas to give nitrogen and hydrogen gases is:

$2NH_3(g)\rightarrow 3H_2(g)+N_2(g)$

From balanced equation, there is 2:3 mol ratio between ammonia and hydrogen. We start with given grams of ammonia and convert them to moles on dividing the grams by molar mass.

In next step the moles of ammonia are multiplied by mol ratio to get the moles of hydrogen which are finally multiplied by it's molar mass to get the grams of hydrogen gas formed.

The set is shown below:

$21.7gNH_3(\frac{1mol NH_3}{17g NH_3})(\frac{3mol H_2}{2mol NH_3})(\frac{2g H_2}{1mol H_2})$

= $3.83g H_2$

So, from the calculations 3.83 g of hydrogen gas are formed by the decomposition of 21.7 g of ammonia gas.