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
So I balanced the equation as 2H3PO4 + 3Ca(OH)2 = 6H2O + Ca3(PO4)2 but now I am stuck

Answer 1

well, you start with 78.5 g of calcium hydroxide. our steps are:

1. convert 78.5g of calcium hydroxide to moles, using molar mass
2. divide moles of calcium hydroxide by 3, since it takes 3 moles of calcium hydroxide to produce 1 mole of calcium phosphate. this gives us the number of moles of calcium phosphate produced
3. multiply that by the molar mass of calcium phosphate to go from moles to grams

Answer 2

So 78.5 g Ca(OH)2 x 74.093 g/mol = 5816.3 mol Ca(OH)2
5816.3 mol / 3 = 1938.8 mol Calcium Phosphate
1938.8 mol x 310.17g/mol = 601,357.6 g
Am I right?

Answer 3

no u are incooorect.
Actually the no of moles are equal to mass divided by the molar mass
so no of moles of Ca(OH)2 would be 78.5g/74.0932gmol^-1
I think u better correct ur answers from that point !

Answer 4

Did u get where u went wrong ?

Answer 5

Okay thank you so
78.5g / 74.093 g/mol = 1.06 mol Ca(OH)2
1.06 mol / 3 = .353 mol Ca3(PO4)2
.353 mol x 310.17 g/mol = 109.49 g
Is this correct now?

Answer 6

Exactly ! U got it right! :)

Answer 7

Is that the answer?

Answer 8

yep !

Answer 9

YAY thank you so much can I ask you another one if you don't mind?

Answer 10

Yeah post it as another question maybe ?

Answer 11

okay i will