For the following reaction, 21.7 grams of diphosphorus pentoxide are allowed to react with with 13.1 grams of water.
What is the maximum amount of phosphoric acid that can be formed in grams?
What is the FORMULA for the limiting reagent?
What amount of the excess reagent remains after the reaction is complete in grams?
This is just another limiting reagent (LR) problem. Follow the guidelines in your later post with BaSO4 et al.
To determine the maximum amount of phosphoric acid that can be formed, the first step is to write and balance the chemical equation for the reaction between diphosphorus pentoxide (P2O5) and water (H2O):
P2O5 + 3H2O → 2H3PO4
The molar mass of P2O5 is 141.94 g/mol, and the molar mass of H3PO4 is 97.99 g/mol.
1. Find the number of moles of diphosphorus pentoxide (P2O5) and water (H2O) using the formula:
Number of moles = Mass / Molar Mass
Number of moles of P2O5 = 21.7 g / 141.94 g/mol = 0.1525 mol
Number of moles of H2O = 13.1 g / 18.02 g/mol = 0.727 mol
2. Next, determine the stoichiometric ratio between P2O5 and H3PO4 using the balanced equation:
According to the equation, 1 mole of P2O5 reacts to form 2 moles of H3PO4.
3. Calculate the maximum number of moles of H3PO4 that can be formed:
For the given amount of P2O5 (0.1525 mol), the number of moles of H3PO4 is:
Number of moles of H3PO4 = 2 * Number of moles of P2O5 = 2 * 0.1525 mol = 0.305 mol
4. Finally, find the mass of H3PO4 formed using the formula:
Mass = Number of moles * Molar Mass
Mass of H3PO4 = 0.305 mol * 97.99 g/mol ≈ 30.01 g
Therefore, the maximum amount of phosphoric acid that can be formed is approximately 30.01 grams.
Now, let's determine the limiting reagent, which is the reactant that will be consumed completely and limits the amount of product that can be formed.
Comparing the stoichiometric ratio between P2O5 and H3PO4, we find that 1 mole of P2O5 reacts to form 2 moles of H3PO4, while the molar ratio between H2O and H3PO4 is 3:2.
1. Calculate the number of moles of H3PO4 that can potentially be formed from the amount of H2O provided:
Number of moles of H3PO4 = (Number of moles of H2O) x (2 mol H3PO4 / 3 mol H2O)
= 0.727 mol x (2 / 3)
= 0.485 mol
2. Compare the number of moles of H3PO4 potentially formed from H2O with the number of moles of H3PO4 formed from P2O5:
Number of moles of H3PO4 from P2O5 = 0.305 mol (calculated earlier)
Since the number of moles of H3PO4 from H2O (0.485 mol) is higher than the number of moles of H3PO4 from P2O5 (0.305 mol), the limiting reagent is P2O5.
Therefore, the formula for the limiting reagent is P2O5.
To find the amount of the excess reagent remaining after the reaction is complete, we need to calculate the amount of the excess reagent initially and subtract the amount that reacted.
1. Calculate the number of moles of the excess reagent (H2O) initially:
Number of moles of H2O initially = 0.727 mol - 0.485 mol (used in the reaction) = 0.242 mol
2. Calculate the mass of the excess reagent remaining using the formula:
Mass = Number of moles * Molar Mass
Mass of excess H2O remaining = 0.242 mol * 18.02 g/mol ≈ 4.36 g
Therefore, the amount of excess reagent (H2O) remaining after the reaction is complete is approximately 4.36 grams.