Question

Question #14025

Answer 1

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.