Question

Consider the reaction. mc015-1.jpg How many grams of N2 are required to produce 100.0 L of NH3 at STP?

Answer 1

The  grams  of N2  that  are   required   to  produce 100.0 l  of  NH3   at  STP

At  stp 1moles = 22.4  l. what  about  100.0 L of NH3

= 100 / 22.4 lx1  moles = 4.46  moles  of NH3

write the reacting  equation

N2+3H2 =2NH3
by use of mole  ratio  between  N2  to NH3  which is  1:2 the moles of N2 =4.46/2 =2.23  moles of N2

mass =  moles  x  molar  mass

=  2.23moles  x 28  g/mol =  62.4 grams