Answer to Question #137960 in General Chemistry for anu

Question

calculate the mass of strontium phosphate that forms when 6.25 grams of strontium chloride react with excess phosphoric acid. hydrochloric acid is the other product of this reaction

Answer 1

m(SrCl₂)= 6.25g

3SrCl₂+ 2H₃PO₄ = Sr₃(PO₄)₂ + 6HCl

n(SrCl₂)= m(SrCl₂)/Mr(SrCl2)= 6.25/(88+35,5*2)=0,039 (mol);

n(Sr₃(PO₄)₂)=n(SrCl₂)/3=0.013 (mol);

m(Sr₃(PO₄)₂)=n(Sr₃(PO₄)₂)*Mr(Sr₃(PO₄)₂)=0.013*454=5.9 (g);

Answer: 5.9 g