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# I am really confused on this problem pleas help me. N2H4(l)+O2(g)(arrow)N2(g)+2H2O(g) a. How many liters of N2 (at STP) form when 1.0 kg N2H4 react

I am really confused on this problem pleas help me.
N2H4(l)+O2(g)(arrow)N2(g)+2H2O(g)
a. How many liters of N2 (at STP) form when 1.0 kg N2H4 react with 1.0 kg O2?
b. How many grams of excess reagent remain after the reaction?1 kg of N2H4 is 1000g/(32g/mol) = 31.25 moles of N2H4. 1 kg of O2 is the same number of moles since it also has a molecular weight of 32 g/mole. The reaction will create an equal number of moles of N2. Each mole (at STP) occupies 22.4 liters. The number of N2 liters formed is therefore 31.25x22.4=
There will be no excess reagent since the mixture is stoichiometric.

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