Question

How many grams of dihydrogen (H) are required to react
completely with 490 g dinitrogen to form ammonia? (atomic mass of
H=1, N = 14)​

Answer 1

Explanation:

The reaction is as follows:  

N2 + 3H2 ⇒ 2NH3

molecular mass of NH3 is 14+3X1 = 17 g  

Hence, 1 mole of NH3 (ammonia)  has mass of 17g.

490 g of NH3 is 490/17 ⇒ 28.82 moles

• 2 parts of NH3 is 28.82 moles
• so,  1 part of NH3 = 28.82/2 ⇒ 14.41 moles  

Therefore, 3 X 14.41 moles of Hydrogen gas is required = 43.23 moles

( As molar mass of H2 = 2 g )

Hence, the mass of Dihydrogen required will be 2 X 43.23=> 86.46 g of Dihydrogen.