Question #137960
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
Expert's answer
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
