Asked by Jaden
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.
Answers
Answered by
DrBob222
See you post above.
Answered by
drwls
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.