Question #14025

1 Answer
Aug 6, 2016

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, #("NH"_4)_3"PO"_4#.

The first reaction involves calcium phosphate, #"Ca"_3("PO"_4)_2# and sulfuric acid, #"H"_2"SO"_4#

#"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#, and ammonia, #"NH"_3#

#"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 #95%#.

This tells you that for every #100# moles of ammonium phosphate that could be produced by the reaction, you only produce #95# moles.

Notice that the first reaction produces #color(red)(2)# moles of phosphoric acid, but that the second reaction only uses #1# mole of phosphoric acid to produce #1# mole of ammonium phosphate.

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 #color(red)(2)# moles of ammonium phosphate for every mole of calcium phosphate that takes part in the first reaction.

Now you're ready to use the #95%# overall yield to say that for every mole of calcium phosphate that takes part in the first reaction, you actually produce

#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, #1# mole of calcium phosphate goes in, #1.9# moles of ammonium phosphate come out. Use the molar mass of ammonium phosphate to convert the given

#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 #1:1.9# mole ratio to find the number of moles of calcium phosphate needed to produce this many moles of ammonium phosphate

#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 #90%# by mass calcium phosphate. This means that for every #"100 t"# of phosphate rock, you get #"90 t"# of calcium phosphate.

Use this ratio to find the mass of phosphate rock that would contain #"1095 t"# of calcium phosphate

#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.