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1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)

If you start with: 5.0 x 10^4 hydrazine (N2H4) "in the tank":

1) How many moles of nitrogen can be created (assuming 100% yield)?
2) How many moles of water can be produced?
2) What mass of dinitrogen tetroxide (N2O4) is needed? What volume of N2O4 is needed? (density = 1.44 g/cm^3)

Can anyone help me on these problems?

4 answers

what is the 5.0x1064 hydrazine in? grams, moles, etc.
sorrry. in grams.
See my response to your later post.
1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)
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