How many molecules of #NH_3# are produced from # 7.52\times10^-4 g# of # H_2#?

1 Answer
Mar 17, 2018

Answer:

I get #1.51*10^20# molecules of ammonia.

Explanation:

We have the familiar Haber process equation...

#N_2(g)+3H_2(g)stackrel(450^@"C")stackrel(200 \ "atm")stackrel(Fe \ "catalyst")rightleftharpoons2NH_3(g)#

Since the mole ratio of #H_2# to #NH_3# is #3:2#, then three moles of hydrogen gas are required to produce two moles of ammonia gas.

We got: #7.52*10^-4 \ "g of" \ H_2#, so let's convert that amount into moles.

#H_2# has a molar mass of #2 \ "g/mol"#. So here, we got

#(7.52*10^-4color(red)cancelcolor(black)"g")/(2color(red)cancelcolor(black)"g""/mol")=3.76*10^-4 \ "mol"#

Since we have #3.76*10^-4# mol of hydrogen, then we are gonna produce:

#3.76*10^-4color(red)cancelcolor(black)("mol" \ H_2)*(2 \ "mol" \ NH_3)/(3color(red)cancelcolor(black)("mol" \ H_2))~~2.51*10^-4 \ "mol" \ NH_3#

We know that one mole of molecules is equal to #6.02*10^23# molecules. So here, we produce

#2.51*10^-4color(red)cancelcolor(black)"mol"*(6.02*10^23 \ "molecules")/(color(red)cancelcolor(black)"mol")~~1.51*10^20 \ "molecules"#