The mass ratio of sodium to fluorine in sodium fluoride is 1.21:1. A sample of sodium fluoride produces 34.5 g of sodium upon decomposition. How much fluorine (in grams) forms?
1 Answer
Explanation:
The idea here is that you need to use the given mass ratio to determine how much fluorine will be produced if the decomposition reaction also produces
A
So based on this information, you know that the mass of fluorine must be smaller than that of sodium.
More specifically, the mass of fluorine will come out to be
#34.5color(red)(cancel(color(black)("g Na"))) * "1 g F"/(1.21color(red)(cancel(color(black)("g Na")))) = color(green)("28.5 g F")#
Alternatively, we can double-check the result by finding the mole ratio that exists between sodium and fluorine in a sample of sodium fluoride.
To do that, use the molar amsses of the two elements. Let's say that you have a sample of sodium fluoride that contains
You know that
#x/y = 1.21/1 = 1.21#
You can say that you have
#(x color(white)(x) color(red)(cancel(color(black)("g"))))/(23.0color(red)(cancel(color(black)("g")))/"mol") = x/23.0"moles of Na"#
#(y color(white)(x) color(red)(cancel(color(black)("g"))))/(19.0color(red)(cancel(color(black)("g")))/"mol") = y/19.0"moles of F"#
The mole ratio will thus be
#x/23.0 * 19.0/y = overbrace(x/y)^(color(blue)(=1.21)) * 19/23 = 1.21 * 19/23 = 0.99956 ~~ 1#
This tells you that you get one mole of fluorine for every one mole of sodium.
The decomposition reaction produced
#34.5color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23/0color(red)(cancel(color(black)("g")))) = "1.5 moles Na"#
Automatically, you know that it also produced
#1.5color(red)(cancel(color(black)("moles"))) * "19.0 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("28.5 g F")#
