Saturday
August 23, 2014

Homework Help: Dr.Bob222

Posted by Brittany on Tuesday, October 27, 2009 at 11:43am.

this is all my work so far on the limiting reagent problem I asked about yesterday about Calcium nitrate and ammonium fluoride....

28.42g/164.1= .17318 mol Ca(NO3)2

29.6g/37.042=.79909 mol (NH4)F

(1mol Ca(NO3)2 /.17318) x (2mol (NH4)F/x)=.34636

(2 mol (NH4)F/.79909)x(1 mol Ca(NO3)2/x)=.399545

so (NH4)F (.34636) is limiting reagent.

now I'm stuck. Not sure what you mean about subtracting moles from initial moles etc.....thanks!

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