Question

1. How many grams of NH3, can we produce when 5.40 g of H2, is reacted with excess N2?

N2 + 3H2(g) → 2 NH3(g)​

Answer 1

30.26 grams of N is produced when 5.40 g of , is reacted with excess nitrogen gas.

Step-by-step explanation:

Answer 2

30.26 grams of N
`H_(3)` is produced when 5.40 g of
`H_(2)`, is reacted with excess nitrogen gas.

Step-by-step explanation:

Data given:

mass of ammonia N
`H_(3)` = ?

mass of Hydrogen gas = 5.40 grams

Balanced chemical reaction:

`N_(2)` + 3
`H_(2)` ⇒2N
`H_(3)`

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 =
`(mass)/(atomic mass of 1 mole)`

putting the values in the equation:

number of moles=
`(5.40)/(2.016)`

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

`(2)/(3)` =
`(x)/(2.67)`

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