What mass of calcium phosphate form by the reaction of 50 mL of 0.2 M H3PO4 with 80 mL of 0.1 M Ca(OH)2?
2 H2PO4 + 3 Ca(OH)2 --> Ca3(PO4)2 + 6 H20
Did you make a typo? H3PO4 and not H2PO4?
2H3PO4 + 3Ca(OH)2 ==> Ca3(PO4)2 + 6H2O
This is a limiting reagent problem. Here is an example (worked) of a limiting reagent problem. Just follow the steps.
Yes I made a typo, sorry. I don't see the example. Can you resend it? Thank you.
To solve this question, you need to use the concept of stoichiometry, which involves determining the relationship between the amounts of reactants and products in a chemical reaction.
First, let's calculate the number of moles of H3PO4 and Ca(OH)2 used in the reaction.
The volume of H3PO4 used is given as 50 mL, and the molarity of H3PO4 is 0.2 M. To calculate the number of moles, we can use the formula:
moles = volume (in liters) x concentration (in moles per liter)
Converting 50 mL to liters:
volume = 50 mL / 1000 = 0.05 L
Now, we can calculate the number of moles of H3PO4:
moles of H3PO4 = 0.05 L x 0.2 mol/L = 0.01 mol
Similarly, for Ca(OH)2:
The volume of Ca(OH)2 used is given as 80 mL, and the molarity of Ca(OH)2 is 0.1 M. Converting 80 mL to liters:
volume = 80 mL / 1000 = 0.08 L
Now, we can calculate the number of moles of Ca(OH)2:
moles of Ca(OH)2 = 0.08 L x 0.1 mol/L = 0.008 mol
According to the balanced equation, the stoichiometric ratio between H3PO4 and Ca(OH)2 is 2:3. This means that 2 moles of H3PO4 react with 3 moles of Ca(OH)2.
Comparing the moles of H3PO4 and Ca(OH)2, we observe that there are fewer moles of Ca(OH)2. Therefore, Ca(OH)2 is the limiting reactant, and we will determine the amount of product (Ca3(PO4)2) formed using the stoichiometric ratio.
From the balanced equation, we know that 3 moles of Ca(OH)2 react to produce 1 mole of Ca3(PO4)2.
So, the number of moles of Ca3(PO4)2 produced is calculated as follows:
moles of Ca3(PO4)2 = (moles of Ca(OH)2) x (1 mole Ca3(PO4)2 / 3 moles Ca(OH)2)
= 0.008 mol x (1/3)
≈ 0.00267 mol
Finally, to determine the mass of Ca3(PO4)2 formed, we need to multiply the number of moles by the molar mass of Ca3(PO4)2.
The molar mass of Ca3(PO4)2 can be calculated by adding the atomic masses of calcium (Ca), phosphorus (P), and oxygen (O) in the compound:
Molar mass of Ca3(PO4)2 = (3 x atomic mass of Ca) + atomic mass of P + (8 x atomic mass of O)
Now, you can substitute the atomic masses and calculate the molar mass of Ca3(PO4)2.
Once you have the molar mass, multiply it by the number of moles calculated earlier to get the mass of Ca3(PO4)2 formed.
I'm sorry. I didn't post it.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html