Answer:
9.3 g of Ca3(PO4)2
Step-by-step explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O
Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:
Molarity of H3PO4 = 0.3 mol/dm³
Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³
Mole of H3PO4 =?
Molarity = mole /Volume
0.3 = mole of H3PO4 /0.2
Cross multiply
Mole of H3PO4 = 0.3 × 0.2
Mole of H3PO4 = 0.06 mole
Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:
3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O
From the balanced equation above,
2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.
Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.
Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.
Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:
Mole of Ca3(PO4)2 = 0.03 mole
Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]
= 120 + 2[31 + 64]
= 120 + 2[95]
= 120 + 190
= 310 g/mol
Mass of Ca3(PO4)2 =?
Mole = mass /Molar mass
0.03 = mass of Ca3(PO4)2 / 310
Cross multiply
Mass of Ca3(PO4)2 = 0.03 × 310
Mass of Ca3(PO4)2 = 9.3 g
Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.