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.
2 answers
See you post above.
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.
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.
3 answers