Consider the following chemical equation.

Na3PO4(aq) + Ca(OH)2(aq) → Ca3(PO4)2(s) + NaOH(aq)
How many grams of calcium hydroxide must react to give 2.81 g of Ca3(PO4)2?

please explain step by step

You first need to balance the equation.

This will give you the ratio of moles of Ca(OH)2 to Ca3(PO4)2.

calculate the molecular mass of
Ca(OH)2 (=Mp) and Ca3(PO4)2 (=Mo)

from the molecular mass (Mp) of Ca3(PO4)2 calculate the number of moles

number of moles = 2.81/Mp

using the ratio from the equation calculate the number of moles (Mo) of Ca(OH)2 needed to form the 2.81/Mp moles

the mass required is then

Mo x molecular mass of Ca(OH)2

To find out how many grams of calcium hydroxide (Ca(OH)2) must react to give 2.81 g of calcium phosphate (Ca3(PO4)2), we need to follow these steps:

Step 1: Determine the molar masses of the compounds involved.
The molar mass of Na3PO4 = 22.99 g/mol (Na) + 3 * (31.0 g/mol (P) + 16.0 g/mol (O)) = 163.94 g/mol.
The molar mass of Ca(OH)2 = 40.08 g/mol (Ca) + 2 * (1.01 g/mol (H) + 16.0 g/mol (O)) = 74.09 g/mol.
The molar mass of Ca3(PO4)2 = 3 * (40.08 g/mol (Ca)) + 2 * (31.0 g/mol (P) + 4 * 16.0 g/mol (O)) = 310.18 g/mol.

Step 2: Calculate the number of moles of Ca3(PO4)2.
To calculate the number of moles, divide the given mass of Ca3(PO4)2 by its molar mass:
Number of moles = Mass / Molar mass = 2.81 g / 310.18 g/mol = 0.00905 mol.

Step 3: Determine the mole ratio between Ca(OH)2 and Ca3(PO4)2 from the balanced chemical equation.
From the balanced equation, we can see that the stoichiometric coefficient of Ca(OH)2 is 1, and the stoichiometric coefficient of Ca3(PO4)2 is 1. Therefore, the mole ratio between Ca(OH)2 and Ca3(PO4)2 is 1:1.

Step 4: Use the mole ratio to calculate the moles of Ca(OH)2.
The moles of Ca(OH)2 can be determined using the mole ratio and the moles of Ca3(PO4)2:
Moles of Ca(OH)2 = Moles of Ca3(PO4)2 = 0.00905 mol.

Step 5: Convert the moles of Ca(OH)2 to grams.
To convert moles to grams, multiply the moles by the molar mass of Ca(OH)2:
Mass = Moles * Molar mass = 0.00905 mol * 74.09 g/mol = 0.672 g.

Therefore, 0.672 grams of calcium hydroxide must react to give 2.81 g of Ca3(PO4)2.