Answer:
See explanation below for answers
Explanation:
This is a stochiometry reaction. LEt's write the overall reaction again:
N₂H₄ + O₂ ---------> N₂ + 2H₂O
This reaction is taking place at Standard temperature and pressure conditions (STP) which are P = 1 atm and T = 273 K. To know the volume of N₂ formed, we need to know first how many moles are formed, and this can be calculated with the reagents and the limiting reagent. Let's calculate the moles first of the reagents:
MM N₂H₄ = 32 g/mol; MM O₂ = 32 g/mol
mol N₂H₄ = 2000 / 32 = 62.5 moles
mol O₂ ? 2100 / 32 = 65.63 moles
Now that we have the moles, we need to apply the stochiometry and calculate the limiting reagent. According to the overall reaction we have a mole ratio of 1:1 between N₂H₄ and O₂, therefore:
1 mole N₂H₄ ---------> 1 mole O₂
62.5 moles ----------> X
X = 62.5 moles of O₂
But we have 65.63 moles, therefore, the limiting reactant is the N₂H₄.
We also have a 1:1 mole ratio with the N₂, so:
moles N₂H₄ = moles N₂ = 62.5 moles
Now that we have the moles, we can calculate the volume with the ideal gas equation:
PV = nRT
V = nRT / P
R: gas constant (0.082 L atm / K mol)
Replacing we have:
v = 62.5 * 0.082 * 273 / 1
V = 1399.13 L of N₂
Now, how many grams of the excess remains?, we know how many moles are reacting so, let's see how much is left:
moles remaining = 65.63 - 62.5 = 3.12 moles
then the mass of oxygen:
m = 3.12 * 32 = 100.16 g of O₂
