Question

How many grams of calcium phosphate can be produced when 78.5 grams of calcium hydroxide reacts with excess phosphoric acid? Unbalanced equation: H3PO4 + Ca(OH)2 → H2O + Ca3(PO4)2 Show, or explain, all of your work along with the final answer

Answer 1

Hey matt could you please help me with this. I know that first we must balance the equation which would be 2H3PO4 + 3Ca(OH)2 --> 6H2O + Ca3(PO4)2 but then I keep getting stuck. I really want to understand this stuff!! :)

Answer 2

Okay so you have the balanced equation of \[2H_{3}PO_{4} + 3Ca(OH)_{2} \rightarrow 6H_{2}O + Ca_{3}(PO_{4})_{2}\]
so now you know you have 78.5 grams of calcium hydroxide and we want the resulting amount of phosphoric acid

Answer 3

Oops I mean we want the resulting amount of calcium phosphate. So we have to do a bunch of conversions

Answer 4

\[78.5gCa(OH)_{2} (\frac{ 1 mol Ca(OH)_{2} }{ 74.093gCa(OH)_{2} })(\frac{ 1 mol Ca_{3}(PO_{4})_{2} }{ 3 mol Ca(OH)_{2} })(\frac{ 310.177 g Ca_{3}(PO_{4})_{2} }{ 1 mol Ca_{3}(PO_{4})_{2} })\]

Answer 5

Hey, so I think I have a good grip on how this should go. I was tutoring a kid earlier today and he had a similar question. This is complicated material so don't feel bad.

You balanced the equation and then you have to set up what is called a stoichiometry equation (which I'm sure you already know about) for each reactant so that you get the same unit for each in order to compare the two.

so for (calcium hydroxide) = 78.5 g ÷ (40.1 + 32 + 2) = 1.0593mol
* Note: I just broke it down in my own sciency way..u can just get the molar mass of calcium hydroxide which is approx 74 g/mol :)

Answer 6

And then for n(calcium phosphate) = 1.0593 ÷ 3 = 0.35312 mol (by mol ratio as seen from equation because when u look at the balanced equation the ratio is 1:3)

∴ ←--- (this menas approx) m(calcium phosphate) = 0.35312 x (40.1 x 3 + 62 + 16 x 8) = 109.57g = 110g 3sf

* NOTE Once again I just broke it down in my science way u can just find that the molar mass of calcium phosphate is approx 310 g/mol and multiply.

Answer 7

Oh so I just do the stoichiometry like I was taught in class for each reactant to compare the two?

Answer 8

Yes That's exactly what you would do! and then once you have compared them, you do one more stoichiometry to find the final mass in grams which would be approx 110 g

Answer 9

Ok Perfect. I just needed to understand how we were supposed to compare the two reactants and the rest I actually got! Thanks for the help

Answer 10

Of course. That's what I'm here for. When do you want to talk next?

Answer 11

if I have an issue on another question I will ask you so it could be soon! This lesson is hard!!