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
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
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