Question

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?

Answer 1

what is the 5.0x1064 hydrazine in? grams, moles, etc.

Answer 2

sorrry. in grams.

Answer 3

See my response to your later post.

Answer 4

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