Balanced Equation
#"3H"_2("g") + "N"_2("g")##rarr##"2NH"_3("g")#
This answer assumes that nitrogen gas is in excess.
One mole of anything is #6.022xx10^23# of anything, including molecules. You need to determine the number of moles of ammonia gas that are produced by the reaction from the given mass of hydrogen gas.
Once you know the moles of ammonia gas produced, you can multiply it by #6.022xx10^23"molecules/mol"#.
Divide the given mass of hydrogen gas by its molar mass #("2.016 g/mol")# by multiplying by its inverse.
#4.10xx10^(-4)color(red)cancel(color(black)("g H"_2))xx(1"mol H"_2)/(2.016color(red)cancel(color(black)("g H"_2)))=2.034xx10^(-4)"mol H"_2#
I am keeping an extra digit in my results to reduce errors caused by rounding. I will round to three sig figs at the end.
Determine the moles of ammonia gas by multiplying the moles of hydrogen gas by the mole ratio between ammonia and hydrogen gas from the balanced equation.
#2.034xx10^(-4)color(red)cancel(color(black)("mol H"_2))xx(2"mol NH"_3)/(3color(red)cancel(color(black)("mol H"_2)))=1.356xx10^(-4)"mol NH"_3#
Multiply the moles of ammonia gas by #("6.022"xx"10"^23" molecules")/(1"mol")#.
#1.356xx10^(-4)("mol NH"_3)xx(6.022xx10^23"molecules NH"_3)/(1color(red)cancel(color(black)("mol NH"_3)))=8.17xx10^19"molecules NH"_3# (rounded to three sig figs)
