Question #14025
1 Answer
Here's what I got.
Explanation:
!! VERY LONG ANSWER !!
Start by writing out the balanced chemical equations that describe the two reactions needed to produce ammonium phosphate,
The first reaction involves calcium phosphate,
#"Ca"_ 3("PO"_ 4) _ (2(s)) + 3"H"_ 2"SO"_ (4(aq)) -> color(red)(2)"H"_ 3"PO"_ (4(aq)) + 3"CaSO"_ (4(s))#
The second reaction involves phosphoric acid,
#"H"_ 3"PO"_ (4(aq)) + 3"NH"_ (3(aq)) -> ("NH"_ 4)_ 3"PO"_ (4(aq))#
Now, here's where the tricky part comes in. You know that overall yield of this process, i.e .the product of the separate yields of the two reactions, is equal to
This tells you that for every
Notice that the first reaction produces
To make the following point easier to understand, rewrite the second equation as
#color(red)(2)"H"_ 3"PO"_ (4(aq)) + 6"NH"_ (3(aq)) -> color(red)(2)("NH"_ 4)_ 3"PO"_ (4(aq))#
Now, if you were to ignore all the other chemical species except calcium phosphate, phosphoric acid, and ammonium phosphate, you could write
#"1 mole Ca"_3("PO"_4)_2 -> color(blue)(cancel(color(black)(color(red)(2)color(white)(a)"moles H"_3"PO"_4)))#
#color(blue)(cancel(color(black)(color(red)(2)color(white)(a)"moles H"_3"PO"_4)))-> color(red)(2)color(white)(a) "moles"color(white)(a)("NH"_4)_3"PO"_4#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#"1 mole Ca"_3("PO"_4)_2 -> color(red)(2)color(white)(a) "moles"color(white)(a)("NH"_4)_3"PO"_4#
This tells you that in theory, your process could produce
Now you're ready to use the
#color(red)(2) color(red)(cancel(color(black)("moles"color(white)(.) ("NH"_4)_3"PO"_4))) * overbrace(("95 moles" color(white)(.)("NH"_4)_3"PO"_4)/(100color(red)(cancel(color(black)("moles"color(white)(.) ("NH"_4)_3"PO"_4)))))^(color(purple)("= 95% overall yield")) = "1.9 moles" color(white)(a)("NH"_4)_3"PO"_4#
So,
#1000 color(red)(cancel(color(black)("t"))) * (10^3color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("t")))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 1 * 10^9"g"#
sample of ammonium phosphate to moles
#1.0 * 10^9 color(red)(cancel(color(black)("g"))) * ("1 mole"color(white)(.)("NH"_4)_3"PO"_4)/(149.087color(red)(cancel(color(black)("g")))) = 6.71 * 10^6"moles"color(white)(a)("NH"_4)_3"PO"_4#
Use the
#6.71 * 10^6 color(red)(cancel(color(black)("moles"color(white)(.)("NH"_4)_3"PO"_4))) * ("1 mole Ca"_3("PO"_4)_2)/(1.9color(red)(cancel(color(black)("moles"color(white)(.)("NH"_4)_3"PO"_4)))) = 3.53 * 10^6"moles Ca"_3("PO"_4)_2#
Use the molar mass of calcium phosphate to convert this to grams
#3.53 * 10^6 color(red)(cancel(color(black)("moles Ca"_3("PO"_4)_2))) * "310.177 g"/(1color(red)(cancel(color(black)("mole Ca"_3("PO"_4)_2)))) = 1095 * 10^6"g"#
This will be equivalent to
#1095 * 10^6 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("kg"))))/(10^3color(red)(cancel(color(black)("g")))) * "1 t"/(10^3color(red)(cancel(color(black)("kg")))) = "1095 t"#
Finally, you know that phosphate rock contains
Use this ratio to find the mass of phosphate rock that would contain
#1095 color(red)(cancel(color(black)("t Ca"_3("PO"_4)_2))) * "100 t phosphate rock"/(90color(red)(cancel(color(black)("t Ca"_3("PO"_4)_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1200 t phosphate rock")color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the mass of ammonium phosphate produced by this chemical process.