30.26 grams of N[tex]H_{3}[/tex] is produced when 5.40 g of [tex]H_{2}[/tex], is reacted with excess nitrogen gas.
Explanation:
Data given:
mass of ammonia N[tex]H_{3}[/tex] = ?
mass of Hydrogen gas = 5.40 grams
Balanced chemical reaction:
[tex]N_{2}[/tex] + 3[tex]H_{2}[/tex] ⇒2N[tex]H_{3}[/tex]
Atomic mass of hydrogen = 2.06 gram/mole
atomic mass of nitrogen = 28.006 grams/mole
atomic mass of ammonia = 17 grams/mole
moles of hydrogen provided
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
putting the values in the equation:
number of moles= [tex]\frac{5.40}{2.016}[/tex]
number of moles = 2.67 moles of hydrogen gas is given
from the chemical reaction:
3 moles of H2 gives 2 moles of NH3
SO, 2.67 moles of H2 will give x moles of NH3
[tex]\frac{2}{3}[/tex] = [tex]\frac{x}{2.67}[/tex]
3x = 5.34
x = 1.78 moles of ammonia
mass of ammonia = number of moles x atomic mass
mass of ammonia = 1.78 x 17
mass of ammonia = 30.26 grams
