If 10.0 L of nitrogen gas reacts with 16.0 L of hydrogen gas, how many liters of ammonia gas (NH3) will be produced?

Volume of ammonia gas : 10.67 L

Given

10.0 L of nitrogen gas

16.0 L of hydrogen gas

Required

Volume of ammonia gas

Solution

Reaction

N₂ + 3H₂ ⇒ 2NH₃

From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles  

Find limiting reactant :

N₂ : 10 : 1 = 10

H₂ : 16 : 3 = 5.3

H₂ as a limiting reactant

So volume of NH₃ based on H₂

= 2/3 x volume H₂

= 2/3 x 16 L

= 10.67 L

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

a) the particles of the chair are strongly held together.