Sample Problem 3.10
(continued)
SOLUTION:
1.00 x 102g N2H4 x
= 3.12 mol N2H4
3.12 mol N2H4 x
= 4.68 mol N2
2.00 x 102g N2O4 x
= 2.17 mol N2O4
2.17 mol N2O4 x
= 6.51 mol N2
N2H4 is the limiting reactant because it produces less product, N2, than does N2O4.
4.68 mol N2 x
= 131g N2
2
4
3
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