Amass of 21.5 grams of calcium hydroxide reacts with an excess of phosphoric acid. what mass of calcium phosphate could be recovered from solution?

The balanced equation for the above reaction is as follows;
3Ca(OH)₂ + 2H₃PO₄ ---> Ca₃(PO₄)₂ + 6H₂O
stoichiometry of Ca(OH)₂ to Ca₃(PO₄)₂ is 3:1
Since Ca(OH)₂ is the limiting reactant, amount of Ca₃(PO₄)₂ formed depends on the amount of Ca(OH)₂ present
Number of Ca(OH)₂ reacted - 21.5 g/74 g/mol = 0.29 mol
Therefore number of Ca₃(PO₄)₂ moles formed - 0.29 mol/3 = 0.0967 mol
Mass of Ca₃(PO₄)₂ formed - 0.0967 mol x 310 g/mol = 30.0 g

Um google the question and answer them.

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.