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
To make 1.50 L of aqueous solution containing 5.00% phosphoric acid by mass, the amount of P4O10 required is:
1.50 L x 1.025 g/mL x 0.05 = 7.875 g
To determine the balanced equation for the reaction of phosphorus (V) oxide, P4O10, dissolving in water to form phosphoric acid, we need to consider the stoichiometry of the reaction.
The reaction can be represented as:
P4O10 + 6 H2O → 4 H3PO4
This equation shows that when 1 mole of P4O10 reacts with 6 moles of water, it forms 4 moles of H3PO4.
Next, we can move on to calculating the number of grams of P4O10 required to make 1.50 L of the solution containing 5.00% phosphoric acid by mass.
To calculate the mass of P4O10, we need to consider the molar mass of P4O10 and the percentage composition of the solution.
The molar mass of P4O10 is:
(4 * atomic mass of P) + (10 * atomic mass of O)
= (4 * 31.0) + (10 * 16.0)
= 124 + 160
= 284 g/mol
Next, we need to find the mass of phosphoric acid in the solution:
Mass of phosphoric acid = Volume of the solution * Density of the solution * Percentage composition of phosphoric acid by mass
= 1.50 L * 1.025 g/mL * 5.00%
= 7.6875 g
Since the stoichiometry of the reaction shows that 1 mole of P4O10 reacts to form 4 moles of H3PO4, we can set up a proportion to calculate the mass of P4O10:
(284 g P4O10) / (4 * 98.00 g H3PO4) = x g P4O10 / 7.6875 g H3PO4
Solving for x, the mass of P4O10 required:
x = (284 g P4O10 * 7.6875 g H3PO4) / (4 * 98.00 g H3PO4)
x ≈ 55.49 g
Therefore, approximately 55.49 grams of P4O10 are required to make 1.50 L of the aqueous solution containing 5.00% phosphoric acid by mass.
To find the balanced equation for the reaction of phosphorus (V) oxide, P4O10, with water to form phosphoric acid, we need to know the individual reactants and the products of the reaction.
The chemical formula for phosphoric acid is H3PO4. The reaction can be written as:
P4O10 + 6H2O → 4H3PO4
This balanced equation shows that 1 mole of P4O10 reacts with 6 moles of water to produce 4 moles of H3PO4.
Now, let's move on to the second part of the question, which asks for the amount of P4O10 required to make a 1.50 L aqueous solution containing 5.00% phosphoric acid by mass with a density of 1.025 g/mL.
First, we need to determine the mass of the phosphoric acid in the solution. Since the solution is 5.00% phosphoric acid by mass, we can calculate the mass of the solution using the density:
Mass of solution = volume of solution × density
Mass of solution = 1.50 L × 1.025 g/mL = 1.5375 kg
To find the mass of phosphoric acid in the solution, we multiply the mass of the solution by the mass percent of phosphoric acid:
Mass of phosphoric acid = mass of solution × mass percent of phosphoric acid
Mass of phosphoric acid = 1.5375 kg × (5.00/100) = 0.076875 kg
Next, we need to calculate the number of moles of phosphoric acid using its molar mass, which is 97.99 g/mol:
Moles of phosphoric acid = mass of phosphoric acid / molar mass of phosphoric acid
Moles of phosphoric acid = 0.076875 kg × (1000 g/kg) / 97.99 g/mol = 0.7846 mol
Finally, we can determine the number of moles of P4O10 required using the balanced equation. From the equation, we know that 4 moles of H3PO4 are produced from 1 mole of P4O10:
Moles of P4O10 = moles of phosphoric acid / (4 mol H3PO4 / 1 mol P4O10)
Moles of P4O10 = 0.7846 mol / (4 mol / 1) = 0.19615 mol
To convert moles to grams, we can use the molar mass of P4O10, which is 283.88 g/mol:
Mass of P4O10 = moles of P4O10 × molar mass of P4O10
Mass of P4O10 = 0.19615 mol × 283.88 g/mol = 55.67 g
Therefore, approximately 55.67 grams of P4O10 are required to make 1.50 L of the aqueous solution containing 5.00% phosphoric acid by mass.