Question

4.5 litres of ammonia at STP was passed over 23.9 g of a heated metallic oxide. 2.3 litres of nitrogen gaswas formed at STP along with 19.05 g of the metal and 5.4 mL of water. Show that these results are inagreement with the law of conservation of mass.​

Answer 1

$\mathrm{NH}_{3}+\mathrm{MO}\rightarrowmathrm{N}_{2}+\mathrm{M}+\mathrm{H}_{2} 0$

As we know at STP 1 mole gas occupy 22.4L  

So, 4.5 L of ammonia (NH3) will occupy = 1/22.4 X 4.5 = 0.2008 mol

Mass of ammonia = 0.2008 X 17 = 3.41 g        [Moles x molecular mass]

Moles of N2 = 1/22.4 X 2.3 = 0.1026 mol

Mass of N2 = 0.1026 X 28 = 2.98 g

Mass of metal oxide which is given in the question = 23.9 g and mass of metal = 19.05 g

Mass of water = 5.4g

Total mass of reactant= 3.41 + 23.9 = 27.31 g

Total mass of product= 2.98 + 19.05 + 5.4 = 27.31 g

We can see that mass of product is equal to the reactant mass so the reaction law of conservation of mass is followed and the results are in agreement with this law.