An excess of AgNO3 reacts with 115.5 mL of an AlCl3 solution to give .215g of AgCl. What is the concentration in moles per liter of the AlCl3 solution?
AlCl3 + 3AgNO3 -> 3AgCl + Al(NO3)3
2007-12-01 12:46:14 · 4 answers · asked by beatrix 1 in Science & Mathematics ➔ Chemistry
Atomic weights: Ag=108 Cl=35.5 AgCl=143.5
Let the AlCl3 solution be called AS
0.215gAgCl/115.5mLAS x 1molAgCl/143.5gAgCl x 1molAlCl3/3molAgCl x 1000mLAS/1LAS = 0.00432 mole/L AlCl3
2007-12-01 13:02:11 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
Turn the .215g of AgCl into moles by dividing the .215 by the molar mass of AgCl. And, then find the moles of AlCl3. After finding moles of AlCl3, divide the moles by .1155 Liters not 115.5 mL. Then, you will get the concentration of the solution.
2007-12-01 20:53:41 · answer #2 · answered by g 1 · 0⤊ 0⤋
you better listen to your chemistry professor because I'm not paying attention to my teacher too, ahaha
2007-12-01 20:59:12 · answer #3 · answered by jayveelim1323 2 · 0⤊ 0⤋
idk im in middle school
2007-12-01 20:53:17 · answer #4 · answered by littleboylitt 2 · 0⤊ 0⤋