How many grams of calcium phosphate (precipitate) could be produced by combining 12.096g of calcium nitrate dissolved in water with 16.498g of rubidium phosphate dissolved in water? Report your answer to 3 decimal places.
This is a limiting reagent problem. I know that because amounts for BOTH reactants are given.
Write and balance the equation.
Convert Ca(NO3)2 to mols. mols = grams/m
molar mass.
Do the same for Rb3PO4.
Using the coefficients in the balanced equation,convert mols of each of the reactants to mols of Ca3(PO4)2. It is likely you will get two values; one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent responsible for that smaller number is the limiting reagent.
Now convert the smaller number to grams. g = mols x molar mass.
Can you explain how to convert mols of each of the reactants to mols of Ca3(PO4)2 more please?
3Ca(NO3)2 + 2Rb3PO4 ==> Ca3(PO4)2 + 6RbNO3
mols Ca(NO3) = about 12.096/164.09 = about 0.0737.
It's a factor. Notice how the units you don't want cancel and the unit you want to convert to remains.
0.0737 mols Ca(NO3)2 x [1 mol Ca3(PO4)3/3 mols Ca(NO3)2] = 0.0737 x (1/3) = ? mols Ca3(PO4)2
Using factor like this you can convert anything to anything. The factors are
[what you have x (mols what you want/mols what you have)] where mols of each are the coefficients in the balanced equation.
To find the number of grams of calcium phosphate (precipitate) that can be produced, we need to determine the limiting reagent. This is the reactant that will be completely consumed first and determines the maximum amount of product that can be formed.
First, we need to find the mole ratio between calcium nitrate and calcium phosphate:
Ca(NO3)2 + 3Rb3PO4 -> 3Ca3(PO4)2 + 6RbNO3
From the balanced equation, we can see that for every 1 mole of calcium nitrate (Ca(NO3)2), we get 3 moles of calcium phosphate (Ca3(PO4)2).
Next, we need to calculate the number of moles for each reactant:
Molar mass of Ca(NO3)2 = (40.08 g/mol (Ca) + (2 × 14.01 g/mol) (N) + (6 × 16.00 g/mol) (O)) = 164.1 g/mol
Number of moles of calcium nitrate = mass / molar mass = 12.096 g / 164.1 g/mol = 0.0736 mol
Molar mass of Rb3PO4 = (3 × 85.47 g/mol (Rb) + 31.00 g/mol (P) + 4 × 16.00 g/mol (O)) = 452.3 g/mol
Number of moles of rubidium phosphate = mass / molar mass = 16.498 g / 452.3 g/mol = 0.0364 mol
Now we compare the mole ratio between calcium nitrate and calcium phosphate:
Moles of calcium nitrate / mole ratio = 0.0736 mol / 1 = 0.0736 mol
Moles of rubidium phosphate / mole ratio = 0.0364 mol / 3 = 0.0121 mol
We can see that the mole ratio between calcium nitrate and calcium phosphate is not limiting the reaction, as there is more calcium nitrate available.
Therefore, the limiting reagent is rubidium phosphate. We can use this information to calculate the maximum amount of calcium phosphate that can be formed.
Moles of rubidium phosphate (Rb3PO4) = 0.0121 mol (limiting reagent)
Using the mole ratio between rubidium phosphate and calcium phosphate (3:1), we can calculate the moles of calcium phosphate that can be produced:
Moles of calcium phosphate = 3 × moles of rubidium phosphate = 3 × 0.0121 mol = 0.0364 mol
Now we need to convert moles of calcium phosphate to grams:
Molar mass of Ca3(PO4)2 = (3 × 40.08 g/mol (Ca) + 2 × 31.00 g/mol (P) + 8 × 16.00 g/mol (O)) = 310.2 g/mol
Mass of calcium phosphate = moles × molar mass = 0.0364 mol × 310.2 g/mol = 11.29 g
Therefore, approximately 11.29 grams of calcium phosphate (precipitate) can be produced when 12.096 grams of calcium nitrate and 16.498 grams of rubidium phosphate are combined.