Posted by Jessica on Sunday, February 6, 2011 at 10:25pm.
I work these by working three "simple" stoichiometry problems and taking the smallest value of moles H2O produced.
For example, for N2H4, it is
kmoles N2H4 = 1200/30.029= 39.96
kmoles H2O = 39.96*(8/2) = 159.85
Do the same for N2O4 and (CH3)2N2H2 and take the smaller kmoles, then
kg H2O = kmoles x 18.015 = ??
The answers you posted are correct.
Remember to round to the correct number of significant figures.
bambi went to see barney
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