Balance the equation and determine how many liter of nitrogen are required to react with 7.3 l of hydrogen to produce ammonia (nitrogen trihydride)?

4.867 L of ammonia

Explanation:

Using Haber's process to form ammonia using Nitrogen and hydrogen, the equation is :

N₂ + 3H₂ → 2NH₃

Here, 3 moles of hydrogen gas gives 2 moles of ammonia.

1 mole of any substance occupies 22.4L at STP

So, 3 x 22.4L of hydrogen gives 2 x 22.4 L of ammonia

Then 7.3 L of hydrogen will give:

=

=

= 4.867 L of ammonia

b is the correct

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g