a 36 grams sample of calcium hydroxide is allowed to react with 40.5 grams of phosphoric acid according to the following reaction.

how many grams of calcium phosphate are produced?

how many moles of calcium phosphate are produced?

how many moles of excess reagent remain?

if 43.5 grams of calcium phosphate are produced what is the % yield?

what reaction?

Kevin,

You can not copy and paste to this site. We do not see what you pasted.

To find the answers to these questions, we will need to follow the steps below:

1. Write and balance the chemical equation for the reaction:
Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O

2. Determine the limiting reactant:
The limiting reactant is the one that will be completely consumed and determines the maximum amount of product that can be formed. To find the limiting reactant, we need to compare the amount of moles of each reactant to their stoichiometric coefficients in the balanced equation.

- Calculate the number of moles for calcium hydroxide:
Moles = mass / molar mass
Molar mass of Ca(OH)2 = 40.08 g/mol (Ca) + 2 * 16.00 g/mol (O) + 2 * 1.01 g/mol (H) = 74.10 g/mol
Moles of Ca(OH)2 = 36 g / 74.10 g/mol

- Calculate the number of moles for phosphoric acid:
Molar mass of H3PO4 = 3 * 1.01 g/mol (H) + 31.00 g/mol (P) + 4 * 16.00 g/mol (O) = 98.00 g/mol
Moles of H3PO4 = 40.5 g / 98.00 g/mol

- Compare the mole ratios:
From the balanced equation, the molar ratio of Ca(OH)2 to H3PO4 is 1:2.
We have calculated the moles of Ca(OH)2 as well as the moles of H3PO4. Divide the moles of each by their stoichiometric coefficients to find the moles per ratio.
Moles of Ca(OH)2 per ratio = Moles of Ca(OH)2
Moles of H3PO4 per ratio = Moles of H3PO4 / 2

- The limiting reactant is the smaller value obtained from the above two calculations.

3. Calculate the moles of calcium phosphate produced:
The stoichiometric ratio between Ca(OH)2 and Ca3(PO4)2 is 1:1 in the balanced equation.
Therefore, the moles of calcium phosphate produced will be equal to the moles of the limiting reactant.

4. Calculate the moles of excess reagent remaining:
The moles of excess reagent remaining can be calculated by subtracting the moles of limiting reactant consumed from the initial moles of the respective reagent.

5. Calculate the percent yield:
Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Theoretical yield is the maximum amount of product obtainable, which is determined by the limiting reactant.
Actual yield refers to the yield obtained in the experiment.

Now, let's go ahead and calculate the answers to these questions using the given data.