Calcium hydroxide and phosphoric acid react to form calcium phosphate and water. Ca(OH)2 +  H3PO4 --> Ca3(PO4)2 + H2OHow many grams of H3PO4 are needed to produce 1,869 grams of Ca3(PO4)2?Write your final answer in two decimal places.Use the following mass numbers:Ca - 40 g/molH - 1 g/molO - 16 g/molP - 31 g/mol

First, we need to calculate the molar mass of each compound involved in the reaction.

1. Calcium hydroxide, Ca(OH)2: Ca: 40 g/mol O: 16 g/mol * 2 = 32 g/mol H: 1 g/mol * 2 = 2 g/mol Total = 40 + 32 + 2 = 74 g/mol

2. Phosphoric acid, H3PO4: H: 1 g/mol * 3 = 3 g/mol P: 31 g/mol O: 16 g/mol * 4 = 64 g/mol Total = 3 + 31 + 64 = 98 g/mol

3. Calcium phosphate, Ca3(PO4)2: Ca: 40 g/mol * 3 = 120 g/mol P: 31 g/mol * 2 = 62 g/mol O: 16 g/mol * 8 = 128 g/mol Total = 120 + 62 + 128 = 310 g/mol

Now, we need to balance the chemical equation. The balanced equation is:

3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6H2O

From the balanced equation, we can see that 2 moles of H3PO4 produce 1 mole of Ca3(PO4)2.

So, to find out how many grams of H3PO4 are needed to produce 1,869 grams of Ca3(PO4)2, we first convert the grams of Ca3(PO4)2 to moles:

1,869 g Ca3(PO4)2 * (1 mol Ca3(PO4)2 / 310 g Ca3(PO4)2) = 6.03 mol Ca3(PO4)2

Then, we use the stoichiometry of the reaction to find out how many moles of H3PO4 are needed:

6.03 mol Ca3(PO4)2 * (2 mol H3PO4 / 1 mol Ca3(PO4)2) = 12.06 mol H3PO4

Finally, we convert the moles of H3PO4 to grams:

12.06 mol H3PO4 * (98 g H3PO4 / 1 mol H3PO4) = 1,181.88 g H3PO4

So, 1,181.88 grams of H3PO4 are needed to produce 1,869 grams of Ca3(PO4)2.