How many grams of Ca3(PO4)2 can be formed from 206 g of calcium nitrate and 140 g of phosphoric acid?

A limiting reagent problem. I know that because amounts are given for BOTH reactants.

1. Write the equation and balance it.
2a. Convert grams Ca(NO3)2 to mols.
mols = grams/molar mass.
2b. Do the same for H3PO4.

3a. Using the coefficients in the balanced equation, convert mols Ca(NO3)2 to mol Ca3(POO4)2.
3b. Do the same for mols H3PO4.
3c. It is likely the two values in 3a and 3b will be agree which means one of them must be wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

4. Using the smaller value convert to grams Ca3(PO4)2. g = mols x molar mass.

How do you balance the equation when you have calcium nitrate and phosphoric acid?

3Ca(NO3)2 + 2H3PO4=>Ca3(PO4)2 + 6HNO3

Thanks!

To find out how many grams of Ca3(PO4)2 can be formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

Let's start by calculating the number of moles of calcium nitrate (Ca(NO3)2) and phosphoric acid (H3PO4) using their molar masses:

Molar mass of Ca(NO3)2:
Ca: 40.08 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3)
Total molar mass: 164.10 g/mol

Number of moles of Ca(NO3)2 = Mass of Ca(NO3)2 / Molar mass of Ca(NO3)2
Number of moles of Ca(NO3)2 = 206 g / 164.10 g/mol

Molar mass of H3PO4:
H: 1.01 g/mol (x3)
P: 30.97 g/mol
O: 16.00 g/mol (x4)
Total molar mass: 98.00 g/mol

Number of moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4
Number of moles of H3PO4 = 140 g / 98.00 g/mol

Next, we need to calculate the stoichiometric ratio of Ca3(PO4)2 to Ca(NO3)2 and H3PO4 from the balanced chemical equation:

3 Ca(NO3)2 + 2 H3PO4 -> Ca3(PO4)2 + 6 HNO3

From the balanced equation, we can see that 3 moles of Ca(NO3)2 react with 2 moles of H3PO4 to form 1 mole of Ca3(PO4)2.

Now, we compare the moles of Ca(NO3)2 and H3PO4 to determine the limiting reactant. The reactant with fewer moles is the limiting reactant:

Moles of Ca(NO3)2 = Calculated moles of Ca(NO3)2
Moles of H3PO4 = Calculated moles of H3PO4

If the moles of Ca(NO3)2 divided by 3 are smaller than the moles of H3PO4 divided by 2, then Ca(NO3)2 is the limiting reactant. Otherwise, H3PO4 is the limiting reactant.

Assuming that Ca(NO3)2 is the limiting reactant, we can calculate the moles of Ca3(PO4)2 formed using the stoichiometric ratio:

Moles of Ca3(PO4)2 = Moles of Ca(NO3)2 / 3

Finally, to calculate the mass of Ca3(PO4)2 formed, we multiply the moles by the molar mass of Ca3(PO4)2:

Mass of Ca3(PO4)2 = Moles of Ca3(PO4)2 * Molar mass of Ca3(PO4)2

Remember to perform the calculations to find the limiting reactant and the corresponding amount of Ca3(PO4)2 formed.