Phosphoric acid is prepared by dissolving phosphorus (V) oxide, P4O10, in water. What is the balanced equation for this reaction? How many grams of P4O10 are required to make 1.50 L of aqueous solution containing 5.00% phosphoric acid by mass? The density of the solution is 1.025 g/mL.
The balanced equation for this reaction is:
P4O10 + 6H2O → 4H3PO4
The number of grams of P4O10 required to make 1.50 L of aqueous solution containing 5.00% phosphoric acid by mass is:
1.50 L x 1.025 g/mL x 0.05 = 7.875 g
The balanced equation for the reaction can be written as:
P4O10 + 6 H2O -> 4 H3PO4
To find the number of grams of P4O10 required, we need to use the given information about the mass percentage and the density of the solution.
1. First, let's calculate the mass of the solution. We can use the density of the solution to convert the volume to mass:
Density = mass / volume
1.025 g/mL = mass / 1.50 L
mass = 1.025 g/mL * 1.50 L = 1.5375 g
2. Now, we can calculate the mass of phosphoric acid in the solution. We know that the solution is 5.00% phosphoric acid by mass, so:
Mass of phosphoric acid = 5.00% * 1.5375 g = 0.076875 g
3. Finally, we can determine the molar mass of P4O10 and calculate the amount of P4O10 required to produce the given mass of phosphoric acid.
Molar mass of P4O10 = (4 * Atomic mass of P) + (10 * Atomic mass of O)
= (4 * 31.0 g/mol) + (10 * 16.0 g/mol)
= 284.0 g/mol
Now, we can use the molar ratio from the balanced equation to find the amount of P4O10 required:
1 mol P4O10 -> 4 mol H3PO4
x g P4O10 -> 0.076875 g H3PO4
Applying the ratio:
x = (0.076875 g H3PO4 * 1 mol P4O10 / 4 mol H3PO4) * (284.0 g/mol)
x ≈ 5.414 g
Therefore, approximately 5.414 grams of P4O10 are required to make 1.50 L of the aqueous solution containing 5.00% phosphoric acid by mass.