Use the following balanced equation.
3Sr(OH)2 + 2H3PO4 --> 6H20 + Sr3(PO4)2
How many grams of Sr3(PO4)2 could be produced by the complete reaction of 200 mL of 0.40 M H3PO4?
See your other post below.
To solve this problem, we need to use the concept of molarity and stoichiometry. Here's how you can calculate the amount of Sr3(PO4)2 produced:
Step 1: Calculate the number of moles of H3PO4.
Molarity (M) = mole/Liter
Given 200 mL of 0.40 M H3PO4,
Volume = 200 mL = 0.2 L (since 1 L = 1000 mL)
Molarity (M) = 0.4 M
Number of moles (n) = Molarity x Volume
n = 0.4 M x 0.2 L
n = 0.08 moles
Step 2: Use the stoichiometry of the balanced equation.
According to the balanced equation, 2 moles of H3PO4 react to produce 1 mole of Sr3(PO4)2.
So, the mole ratio between H3PO4 and Sr3(PO4)2 is 2:1.
Step 3: Calculate the moles of Sr3(PO4)2.
Number of moles of Sr3(PO4)2 = 0.08 moles x (1 mole Sr3(PO4)2 / 2 moles H3PO4)
Number of moles of Sr3(PO4)2 = 0.04 moles
Step 4: Calculate the mass of Sr3(PO4)2.
To calculate the mass of Sr3(PO4)2, we need to know its molar mass.
Sr3(PO4)2 consists of strontium (Sr) and phosphate (PO4) ions.
Molar mass of strontium (Sr) = 87.62 g/mol
Molar mass of phosphate (PO4) = 94.97 g/mol
Molar mass of Sr3(PO4)2 = (3 x Molar mass of Sr) + (2 x Molar mass of PO4)
Molar mass of Sr3(PO4)2 = (3 x 87.62 g/mol) + (2 x 94.97 g/mol)
Molar mass of Sr3(PO4)2 = 525.36 g/mol + 189.94 g/mol
Molar mass of Sr3(PO4)2 ≈ 715.3 g/mol
Mass of Sr3(PO4)2 = Number of moles x Molar mass
Mass of Sr3(PO4)2 = 0.04 moles x 715.3 g/mol
Mass of Sr3(PO4)2 ≈ 28.612 g
Therefore, approximately 28.612 grams of Sr3(PO4)2 could be produced by the complete reaction of 200 mL of 0.40 M H3PO4.