If you have 225 grams of Ca3(PO4)2, and 185g H3PO4; calculate which is the limiting reagent, which the XS reagent, and how much product you will produce from the limiting reagent (in grams)

Ca(PO4)2 + 4H3PO4 = 3Ca(H2PO4)2

Emily, what do you not understand about limiting reagent problems. Actually they are just two stoichometry problems put together.

To determine the limiting reagent and excess reagent, we need to calculate the number of moles for each reactant. Then, we can compare the mole ratios between the reactants to identify the limiting and excess reagents.

First, let's calculate the moles of each reactant:

1. Calculate the moles of Ca3(PO4)2:
Moles of Ca3(PO4)2 = mass (in grams) / molar mass (in grams/mole)
Molar mass of Ca3(PO4)2 = (40.1 g/mol * 3) + (31 g/mol * 2) + (16 g/mol * 8)
= 120.3 g/mol + 62 g/mol + 128 g/mol
= 310.3 g/mol
Moles of Ca3(PO4)2 = 225 g / 310.3 g/mol

2. Calculate the moles of H3PO4:
Molar mass of H3PO4 = (1 g/mol * 3) + (16 g/mol) + (1 g/mol * 4)
= 3 g/mol + 16 g/mol + 4 g/mol
= 23 g/mol
Moles of H3PO4 = 185 g / 98 g/mol

Now, let's compare the mole ratios of the reactants:

According to the balanced equation, the mole ratio between Ca3(PO4)2 and H3PO4 is 1:4. That means for every 1 mole of Ca3(PO4)2, 4 moles of H3PO4 are required.

3. Determine the limiting reagent:
By comparing the mole ratios, we can determine which reactant is present in the fewest moles. The limiting reagent is the one that produces the smallest amount of product.
Calculate the moles of product that can be formed from Ca3(PO4)2:
Moles of Ca(H2PO4)2 = Moles of Ca3(PO4)2 * (3 mol Ca(H2PO4)2 / 1 mol Ca3(PO4)2)

Calculate the moles of product that can be formed from H3PO4:
Moles of Ca(H2PO4)2 = Moles of H3PO4 * (3 mol Ca(H2PO4)2 / 4 mol H3PO4)

The smaller value between the two calculations above will determine the amount of product that can be formed.

4. Determine the excess reagent:
The excess reagent is the reactant that is not entirely consumed in the reaction.
We can calculate the moles of the excess reagent by subtracting the moles used in the reaction from the initial moles of the respective reagent.

Finally, let's calculate the grams of product formed from the limiting reagent:

5. Calculate the mass of Ca(H2PO4)2 produced from the limiting reagent:
Mass of Ca(H2PO4)2 = Moles of Ca(H2PO4)2 * Molar mass of Ca(H2PO4)2

By following these steps, you should be able to determine the limiting reagent, excess reagent, and the mass of product produced from the limiting reagent.