Asked by Alekya
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.
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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