How many grams of calcium phosphate (precipitate) could be produced by combining 12.443g of calcium nitrate dissolved in water with 16.083g of rubidium phosphate dissolved in water? Report your answer to 3 decimal places.
See you post above.
To determine how many grams of calcium phosphate could be produced, we need to use stoichiometry, which involves the balanced chemical equation for the reaction between calcium nitrate and rubidium phosphate.
The balanced chemical equation for the reaction is:
3Ca(NO3)2 + 2Rb3PO4 --> Ca3(PO4)2 + 6NO3 + 6RbNO3
From the equation, we can see that 3 moles of calcium nitrate react with 2 moles of rubidium phosphate to produce 1 mole of calcium phosphate.
To find the moles of calcium nitrate and rubidium phosphate given their masses, we need to divide the mass of each compound by their respective molar masses.
Calcium nitrate (Ca(NO3)2) has a molar mass of:
Ca: 40.08 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (three oxygen atoms)
So, the molar mass of calcium nitrate is:
40.08 g/mol + (14.01 g/mol + 16.00 g/mol * 3) * 2 = 164.10 g/mol
Dividing the given mass of calcium nitrate (12.443 g) by its molar mass, we can calculate the moles of calcium nitrate:
12.443 g / 164.10 g/mol ≈ 0.0757 mol (rounded to four decimal places)
Rubidium phosphate (Rb3PO4) has a molar mass of:
Rb: 85.47 g/mol
P: 30.97 g/mol
O: 16.00 g/mol (four oxygen atoms)
So, the molar mass of rubidium phosphate is:
(85.47 g/mol * 3) + 30.97 g/mol + (16.00 g/mol * 4) = 398.34 g/mol
Dividing the given mass of rubidium phosphate (16.083 g) by its molar mass, we can calculate the moles of rubidium phosphate:
16.083 g / 398.34 g/mol ≈ 0.0404 mol (rounded to four decimal places)
Based on the balanced chemical equation, the mole ratio between calcium nitrate and calcium phosphate is 3:1. Therefore, the number of moles of calcium phosphate formed is equal to the number of moles of calcium nitrate.
So, we have 0.0757 mol of calcium phosphate.
To find the mass of calcium phosphate produced, we multiply the number of moles by the molar mass of calcium phosphate.
The molar mass of calcium phosphate (Ca3(PO4)2) is:
(40.08 g/mol * 3) + (30.97 g/mol + (16.00 g/mol * 4) * 2) = 310.18 g/mol
The mass of calcium phosphate produced is:
0.0757 mol * 310.18 g/mol ≈ 23.472 g (rounded to three decimal places)
Therefore, approximately 23.472 grams of calcium phosphate precipitate can be produced by combining 12.443 grams of calcium nitrate and 16.083 grams of rubidium phosphate.
The balanced reaction is
2 Rb3PO4 + 3 Ca(NO3)2
-> Ca3(PO4)2 + 6 RbNO3
Calculate the number of moles of each reactant.
The molar mass of Rb3PO4 is 351.41 g/mol
The molar mass of Ca(NO3)2 is 164.10 g/mol
You are reacting 0.07583 moles of the Ca salt with 0.04577 moles of the Rb salt.
One of the reactants (rubidium phosphate) appears to be limiting.
Take it from there.