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.
700 liter of N
7.0*102 l n
700 L boiiii
To find the number of liters of N2 formed when 1.0 kg N2H4 reacts with 1.0 kg O2, we first need to calculate the number of moles of N2H4 and O2.
Given:
Mass of N2H4 = 1.0 kg
Molar mass of N2H4 = 32 g/mol
Since 1 kg is equal to 1000 g, the number of moles of N2H4 can be calculated as follows:
Number of moles = Mass / Molar mass = 1000 g / 32 g/mol = 31.25 moles
Similarly, we can find the number of moles of O2 since its molar mass is also 32 g/mol:
Number of moles = Mass / Molar mass = 1000 g / 32 g/mol = 31.25 moles
Now, according to the balanced equation, 1 mole of N2H4 produces 1 mole of N2.
Therefore, the number of moles of N2 formed is equal to the number of moles of N2H4 reacting.
Using Avogadro's Law, we can convert the number of moles of N2 to liters at STP (Standard Temperature and Pressure), where 1 mole of any gas occupies 22.4 liters.
Number of liters of N2 formed = 31.25 moles x 22.4 liters/mole = 700 liters
Therefore, 700 liters of N2 (at STP) will be formed when 1.0 kg N2H4 reacts with 1.0 kg O2.
As for the second part of the problem, since the reaction is stoichiometric, there will be no excess reagent leftover after the reaction. Stoichiometric means that the reactants are in exactly the right proportions to react completely without any excess left over.
Therefore, in this reaction, no grams of excess reagent will remain after the reaction.