Question

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: N2(g) + 3H2(g) → 2NH3(g)

1. What is the maximum mass of ammonia that can be produced from a mixture of 216.5 g of N2 and 51.40 g of H2? g

2. Which element would be completely consumed? (enter nitrogen or hydrogen)

3. What mass of the starting material would remain unreacted?

Answer 1

According to reaction 1 mole of nitrogen react with 3 mole of hydrogen to produce 2 mole of ammonia

molar mass of N₂ =28.0134 g/mol and

molar mass of H₂ = 2.01 g/mol then molar mass of 3H₂ =6.03 g/mol

28.0134 gm of N₂ react with 6.03 gm of H₂ then 216.5 gm N₂ require

216.56.03/28.0134 =46.60 gm of H₂

N₂ react completly and 51.40 - 46.60 = 4.8 gm H₂ remain unreacted

molar mass of NH₃ = 17.031 gm/mol then 2NH₃ = 34.062 gm

1 mole of nitrogen react with 3 mole of hydrogen to produce 2 mole of ammonia and N₂ react comletly that mean

28.0134 gm N₂ produce 34.062 gm NH₃ then 216.5 gm produce how many gm NH₃ calculation are as follow

gm of NH₃ = 216.534.062/28.0134 = 263.246 gm

maximum mass of ammonia = 263.246 gm

2) N₂ consumed completely

3)

molar mass of N₂ =28.0134 g/mol and

molar mass of H₂ = 2.01 g/mol then molar mass of 3H₂ =6.03 g/mol

28.0134 gm of N₂ react with 6.03 gm of H₂ then 216.5 gm N₂ require

216.56.03/28.0134 =46.60 gm of H₂

N₂ react completly and 51.40 - 46.60 = 4.8 gm H₂ remain unreacted