Question #c35b1
1 Answer
Explanation:
Start by taking a look at the balanced chemical equation for this reaction
#3"H"_text(2(g]) + "N"_text(2(g]) -> color(red)(2)"NH"_text(3(g])#
Now, you know that the reaction produced
So, if the reaction produced
#4color(red)(cancel(color(black)("moles NH"_3))) * ("3 moles H"_2)/(color(red)(2)color(red)(cancel(color(black)("moles NH"_3)))) = "6 moles H"_2#
By the same logic, the
#4color(red)(cancel(color(black)("moles NH"_3))) * ("1 mole N"_2)/(color(red)(2)color(red)(cancel(color(black)("moles NH"_3)))) = "2 moles N"_2#
To test the result, check to see if the number of moles of each reactant satisfy the
As you can see,
Since you started with
#n_(N_2) = 2.5 - 2 = color(green)("0.5 moles N"_2)#
Likewise, the number of moles of hydrogen that remain will be
#n_(H_2) = 10 - 6 = color(green)("4 moles H"_2)#