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Answer 1

Answer #1

N2O4(l) + 2N2H4(l) ---------------> 3N2(g) + 4H2O(g)

1 mole of N2O4 react with excess N2H4 to gives 3 moles of N2

92g of N2O4 react with excess N2H4 to gives 3*28g of N2

2550g of N2O4 react with excess N2H4 to gives = 3*28*2550/92 =2328.3g pf N2

= 2.3*10^3 g of N2 >>>>>>>answer

Answer 2

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