Question #d111c
1 Answer
Explanation:
Start by taking a look at the balanced chemical equation that describes this redox reaction
#color(red)(4)"NH"_ (3(g)) + 3"O"_ (2(g)) -> 2"N"_ (2(g)) + color(blue)(6)"H"_ 2"O"_((l))#
Since no mention of the number of molecules of oxygen gas available for this reaction was made in the problem, you can assume that oxygen gas is in excess.
This means that all the moles of ammonia,
As you can see, the reaction produces
Now, a mole is simply a very large collection of molecules. More specifically, in order to have one mole of a substance, you need to have
#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 mole" = 6.022 * 10^(23)"molecules")color(white)(a/a)|))) -># Avogadro's number
This means that you can convert the mole ratio to a molecule ratio by using Avogadro's number as a conversion factor. So, instead of
#color(blue)(6)color(white)(a) "moles H"_2"O" = color(blue)(6) xx 6.022 * 10^(23)"molecules H"_2"O"#
#color(red)(4)color(white)(a)"moles NH"_3 = color(red)(4) xx 6.022 * 10^(23)"molecules NH"_3#
This will be equivalent to
#(color(blue)(6)color(white)(a)"moles H"_2"O")/(color(red)(4)color(white)(a)"moles NH"_3) = (color(blue)(6) xx color(black)(cancel(color(black)(6.022 * 10^(23))))"molecules H"_2"O")/(color(red)(4) xx color(black)(cancel(color(black)(6.022 * 10^(23))))"molecules NH"_3)#
As you can see, a mole ratio will always be equivalent to a molecule ratio because of the fact that a mole is simply a set number of molecules
#color(purple)(|bar(ul(color(white)(a/a)color(black)((color(blue)(6)color(white)(a)"moles H"_2"O")/(color(red)(4)color(white)(a)"moles NH"_3) = (color(blue)(6)color(white)(a)"molecules H"_2"O")/(color(red)(4)color(white)(a)"molecules NH"_3)color(white)(a/a)|)))#
So, if the reaction uses up
#8.02 * 10^(23) color(red)(cancel(color(black)("molec. NH"_3))) * (color(blue)(6)color(white)(a)"molec. H"_2"O")/(color(red)(4)color(red)(cancel(color(black)("molec. NH"_3)))) = color(green)(|bar(ul(color(white)(a/a)1.20 * 10^(24)"molec. H"_2"O"color(white)(a/a)|)))#
The answer is rounded to three sig figs.