Kirsten is in her lab, and reacts two unknown gases (A(g) and B(g)) to get nitrogen dioxide (NO2). Using her data, she is able to write the following balanced equation:
2A(g) + B(g) --> 2NO2(g)
If she started with 8 L of A(g) and 3 L of B(g), how many liters of NO2(g) would she get? [Assume everything is at STP.]
2 L
3 L
4 L
6 L
8 L
When gases react to produce a gas, we need not go through the mol step; i.e., we can use liters as if liters were mols.
You can see this is a limiting reagent problem. It takes twice as much A as B; therefore, it would take 4 L of B to react with 8 L of A and we don't have 4 L of B. So I guessed wrong. 3 L of B will require 6 liters of A and we have more than enough. So B is the limiting reagent and all of B reacts. How much of the product will be formed? And you have your answer.
To find out how many liters of NO2(g) Kirsten would get, we need to use the stoichiometry of the balanced equation.
According to the balanced equation, 2 moles of A(g) react with 1 mole of B(g) to produce 2 moles of NO2(g). Therefore, the ratio of the reactants to the product is 2:1:2.
To determine the number of moles of NO2(g) produced, we need to calculate the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
First, let's calculate the number of moles of A(g) and B(g) using the ideal gas law equation:
n = PV/RT
Where:
n = number of moles
P = pressure (STP, 1 atm)
V = volume in liters
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (STP, 273 K)
For A(g):
n(A) = (1 atm × 8 L) / (0.0821 L·atm/mol·K × 273 K) = 0.366 moles
For B(g):
n(B) = (1 atm × 3 L) / (0.0821 L·atm/mol·K × 273 K) = 0.137 moles
From the balanced equation, we see that the mole ratio of A(g) to B(g) is 2:1. Therefore, the 0.137 moles of B(g) would require 0.274 moles of A(g) to react completely.
However, Kirsten only has 0.366 moles of A(g), which is more than enough to react with 0.137 moles of B(g). This means B(g) is the limiting reactant, and A(g) is in excess.
Using the mole ratio from the balanced equation, we know that 2 moles of NO2(g) are produced for every 1 mole of B(g) reacted. Therefore, the number of moles of NO2(g) produced is equal to the number of moles of B(g).
So, Kirsten would produce 0.137 moles of NO2(g).
Finally, we can convert the moles of NO2(g) to liters using the ideal gas law equation:
V = nRT/P
V(NO2) = (0.137 moles × 0.0821 L·atm/mol·K × 273 K) / 1 atm = 2 L
Therefore, Kirsten would produce 2 liters of NO2(g).
The correct answer is: 2 L.