When 3.50 LITERS of N2H4react with 3.5 LITERS of N2O4, how many LITERS of N2will be produced at STP
I did this down below :)
Hey, I am just an old retired physicist. I do not work fast.
Now check my arithmetic carefully because my brain is a bit fuzzy.
Once again :
3NO2+ 1H2O ------> 1NO + 2HNO3
How many moles of NO2 in 9 liters at stp?
9 L * 1 mol / 22.4 L = 0.401 mols
If that is liquid water it is about (7 /1000) meters^3
water mass is about 1000 kg/m^3 so about 7 kg = 7000 grams
mol H2O is 2+16 = 18 g /mol
so we have an excess of water, use 0.301 mols NO2
for every 3 liters of NO2 we get ONE liter of NO
so Three
The same way.
2 N 2 H 4 + N 2 O 4 → 3 N 2 + 4 H 2 O
note you need 2 liters of the first for 1 of the second
so N2H4 governs the problem
we get 3 liters of N2 for every 2 liters of N2H4
so
(3/2)(3.5) Liters
To determine the volume of N2 produced at STP when 3.50 liters of N2H4 reacts with 3.5 liters of N2O4, we first need to balance the chemical equation for the reaction.
The balanced chemical equation is:
N2H4 + N2O4 → 2N2 + 4H2O
From the balanced equation, we can see that for every 1 mole of N2H4 reacting, it produces 2 moles of N2.
To solve the problem, we need to follow these steps:
Step 1: Convert the given volumes of N2H4 and N2O4 to moles.
We can use the ideal gas law, where n (moles) = V (volume in liters) / Vm (molar volume at STP, which is 22.4 liters/mol).
For N2H4:
n(N2H4) = 3.50 L / 22.4 L/mol
For N2O4:
n(N2O4) = 3.5 L / 22.4 L/mol
Step 2: Determine the limiting reactant.
To find the limiting reactant, compare the moles of N2H4 and N2O4 and identify the reactant that has the smallest number of moles. The reactant with the smallest number of moles determines the maximum amount of product that can be formed.
Step 3: Determine the moles of N2 produced using the stoichiometry of the balanced equation.
Since 1 mole of N2H4 produces 2 moles of N2, we can multiply the moles of N2H4 by 2 to get the moles of N2.
moles of N2 = 2 * n(N2H4)
Step 4: Convert the moles of N2 to liters.
Again, using the ideal gas law, where V (volume in liters) = n (moles) * Vm (molar volume at STP, which is 22.4 liters/mol), we can convert the moles of N2 to liters.
volume of N2 = moles of N2 * 22.4 L/mol
By following these steps and plugging in the given values, you can find the volume of N2 produced at STP.