posted by .

the question reads:

a) balance the equation:
H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O
b)what mass of each product results if 750 mL of 6.00 M H3PO4 reacts according to the equation?

i balanced it with:
2H3PO4 + 3Ca(OH)2 --> Ca3(PO4)2 + 6H2O

but i'm not sure where to go from there


A solution is made by dissolving 26.42 g of (NH4)2SO4 in enough H2O to make 50.00 mL of solution.

a) what is the molar mass of (NH4)2SO4?
b) what are the products of the solution?
c) how many milliliters of solution are needed?

i got:
a) 196g/mol
b) (NH4)2SO4 + H2O --> NH4O + SO4H
c) 2.7 M

i'm sure a and c is right.. just unsure about b

sorry for the long post, thanks !

  • chemistry -

    H3PO4 equation is balanced.
    Convert what you have to mols. M x L = mols.
    Convert mols of what you have to mols of what you want.
    Convert mols of what you want to grams.
    grams = mols x molar mass.

    For the second question, which is incomplete,
    a is 132.141 but check my work since it doesn't agree with yours.
    b. You haven't reacted it with anything but water, It will ionize into NH4^+ and SO4^=
    c. How many mL are needed for what??

  • chemistry -

    What I would first is to find the amount of mols of H3PO4. Since you know that 1 M means there is 1 mol of solution in every 1 liter, you can figure out the moles of the solution. The solution is 6M, so that means that to find the moles of solute in the solution you multiply .750 (750 milliliters converted to liters) by 6. I think you did something similar, but .0075 L makes 7.5 milliliters. This will give you 4.5 mols of H3PO4.

    Now you can use stoichiometry to find the grams of Ca3(PO4)2. The balanced equation states that it takes 2 mols of H3PO4 to get 1 mol of Ca3(PO4)2:

    4.5 mol H3PO4 * 1 mol Ca3(PO4)2 / 2 mol H3PO4 = 2.25 mol Ca3(PO4)2

    Find the molar mass of Ca3(PO4)2 to convert the mols to grams:
    3*40 + 2*31 + 8*16 = 310 grams / mol

    2.25 mol Ca3(PO4)2 * 310 grams Ca3(PO4)2 / 1 mol Ca3(PO4)2 = 697.5 grams Ca3(PO4)2

    You can use the same process to find the mass of water.

    4.5 mol H3PO4 * 6 mol H2O / 2 mol H3PO4 = 13.5 mol H2O

    13.5 mol H2O * 18 g H2O / 1 mol H2O = 243 g H2O

    Hope that helped! NB not my work. its from another website

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    Wow, ok, that helps a LOT!! I guess you can't very easily balance an equation that can never happen! :) Thanks a lot! Just to make things clear, wouldn't it have to be Mg3(PO4)2, from balancing the charges?
  2. Chemistry

    Balance the following chemical equation. __Fe(OH)2 (s) + __H3PO4 (aq) -> __Fe3(PO4)2 (s) + __ H2O (l) Would it be 3 Fe(OH)2 (s) + 2 H3PO4 (aq) -> 1 Fe3(PO4)2 (s) + 6 H2O (l) ?
  3. Chemistry

    1. Ca(OH)2(aq) + H3PO4(aq) à Ca3(PO4)2(aq) + H2O(l) a. If you have 13.7mol of Ca(OH)2, how many grams of H2O are produced?
  4. Chemistry

    1. Ca(OH)2 + H3PO4 ---> H2O + Ca3(PO4)2 (A) How many moles of H3PO4 are needed to make 15g of H2O?
  5. Chemistry

    Use the following balanced equation. 3Sr(OH)2 + 2H3PO4 --> 6H20 + Sr3(PO4)2 How many mL of 0.10 M Sr(OH)2 are required to react with 30 mL of 0.30 M H3PO4?
  6. chemistry

    Use the following balanced equation. 3Sr(OH)2 + 2H3PO4 --> 6H20 + Sr3(PO4)2 How many grams of Sr3(PO4)2 could be produced by the complete reaction of 200 mL of 0.40 M H3PO4?
  7. chemistry

    Use the following balanced equation. 3Sr(OH)2 + 2H3PO4 --> 6H20 + Sr3(PO4)2 20 mL of H3PO4 is required to react with 40 mL of a 0.15 M Sr(OH)2 solution. What is the molarity of the H3PO4 solution?
  8. chem

    Consider the following neutralisation reaction: 2 H3PO4(aq) + 3 Ca(OH)2 (aq) → Ca3(PO4)2 + 6 H2O Determine the volume of 0.75 M H3PO4 necessary to neutralise 35 ml of 0.45 M Ca(OH)2
  9. chemistry

    H3PO4+3Ca(OH)2=Ca3(PO4)2=6H2O If 44.6g of Ca3(PO4)2 are formed in the above reaction, how many moles of H3PO4 must have been consumed?
  10. chemistry

    What is the chemical equation that describes the complete neutralization of H3PO4 by NaOH?

More Similar Questions