Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: meee45 on December 11, 2005, 02:30:19 PM
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wats the limiting reactant for this quest...
2Al + 3CuCl3 -------------> 3CU + 2AlCl3
Aluminum - 0.25 g
copper (II) chloride - 0.51 g
i keep getting CuCl2 but the way im doing it is wrong....am i suppose tp calculate the amount of Cu produced or AlCl3? som1 plz help
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1.)Balance the reaction equation and make sure that you use the correct formulas
2.)calculate how many moles 0.25g of Al and 0.51g of CuCl2 are
Show your results and you'll get other instructions!
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a). its already balanced
b). # mol. Al = 0.25g / 26.98 g/mol
= 0.009mol
# mol CuCl3 = 0.51g / 169.9 g/mol
= 0.003 mol
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a) your reaction is not balanced properly
Copper (II) chloride would be CuCl2 as Alberto mentioned.
CuCl3 would be Copper (III) chloride. So which one was your question? Because you said you were given the amount of copper (II) chloride, but then you switched your conversion for copper (III) chloride
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soo sorry i typed it wrong
its actually 2Al + 3CuCl2 ------>3Cu + 2AlCl3
so yea its balanced. but my quest. is that am i suppose to calculate the amount of Cu produced from each reactant or the amount of AlCl3 produced.
i hav the rite ans. but its jus that my cal. is wrong...
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either one should give you the right answer. The gfm of CuCl2 is 134.45 so that may be why you are having trouble. If you calculate the Cu produced from each of the reactant, you can find the limiting reactant by comparing the yield. The reactant that produces less Cu is your limiting reactant.
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thank you sooooooooooooo much!
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Red means in excess 8)
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lol thats so cool. thank u for your help. ;D ;)