Question

According to the equation below, how many moles of Ca(OH)2 are required to react with 1.36 mol H3PO4 to produce Ca3(PO4)2? 3Ca(OH)2+2H3PO4⟶Ca3(PO4)2+6H2O

Answer 1

Answer: The amount of calcium hydroxide needed to react is 2.04 moles

Explanation:

We are given:

Moles of phosphoric acid = 1.36 moles

For the given chemical equation:

$3Ca(OH)_2+2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O$

By Stoichiometry of the reaction:

2 moles of phosphoric acid reacts with 3 moles of calcium hydroxide

So, 1.36 moles of phosphoric acid will react with = $\frac{3}{2}\times 1.36=2.04mol$ of calcium hydroxide

Hence, the amount of calcium hydroxide needed to react is 2.04 moles

Answer 2

The number of mole of Ca(OH)₂ required to react with 1.36 mole of H₃PO₄ is 2.04 moles

The balanced equation for the reaction is given below:

3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

3 moles of Ca(OH)₂ reacted with 2 moles of H₃PO₄

• With the above information in mind, we can obtain the number of mole of Ca(OH)₂ required to react with 1.36 mole of H₃PO₄. This can be obtained as follow:

From the balanced equation above,

3 moles of Ca(OH)₂ reacted with 2 moles of H₃PO₄

Therefore,

X mole of Ca(OH)₂ will react with 1.36 moles of H₃PO₄ i.e

X mole of Ca(OH)₂ = (3 × 1.36)/2

X mole of Ca(OH)₂ = 2.04 moles

Thus, the number of mole of Ca(OH)₂ required to react with 1.36 mole of H₃PO₄ is 2.04 moles

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