How many liters of excess reactant would be left when 3.00 liters of nitrogen monoxide is mixed with 2.00 liters of oxygen and allowed to react at 695 torr and 27 degrees Celsius to produce nitrogen dioxide

Question

How many liters of excess reactant would be left when 3.00 liters of nitrogen monoxide is mixed with 2.00 liters of oxygen and allowed to react at 695 torr and 27 degrees Celsius to produce nitrogen dioxide

Answer 1

Start from the equation

2NO + O2 => 2NO2

so 2 moles + 1 mole gives 2 moles

or

2 volumes + 1 volume gives 2 volumes

Now decide which of the starting materials in excess

Answer 2

To determine the excess reactant and the amount of it left, we need to compare the stoichiometric amounts of reactants. Let's start by writing and balancing the chemical equation for the reaction:

2 NO (nitrogen monoxide) + O2 (oxygen) → 2 NO2 (nitrogen dioxide)

Next, we need to convert the given volumes of gases at 27 degrees Celsius to the number of moles using the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.08206 L·atm/mol·K)
T = temperature (in Kelvin)

Converting the temperatures:
27 degrees Celsius = 27 + 273 = 300 Kelvin

Now we can calculate the number of moles for each gas:

For nitrogen monoxide (NO):
P(NO) = 695 torr = 695/760 atm
V(NO) = 3.00 L
T = 300 K

n(NO) = (P(NO) * V(NO)) / (R * T)
n(NO) = (695/760 atm * 3.00 L) / (0.08206 L·atm/mol·K * 300 K)
n(NO) ≈ 0.256 mol

For oxygen (O2):
P(O2) = 695 torr = 695/760 atm
V(O2) = 2.00 L
T = 300 K

n(O2) = (P(O2) * V(O2)) / (R * T)
n(O2) = (695/760 atm * 2.00 L) / (0.08206 L·atm/mol·K * 300 K)
n(O2) ≈ 0.172 mol

Now, let's compare the moles of the reactants to find the limiting reactant:

According to the balanced chemical equation, we need 2 moles of NO for every 1 mole of O2. So, the stoichiometric ratio is 2:1 NO to O2.

Using the stoichiometry, we can determine the number of moles of NO needed for the given moles of O2:

moles of NO needed = 2 * moles of O2
moles of NO needed = 2 * 0.172 mol
moles of NO needed ≈ 0.344 mol

Since we have less NO (0.256 mol) than what is needed (0.344 mol), NO is the limiting reactant, and O2 is the excess reactant.

To find the excess reactant remaining, we'll use the stoichiometry again:

moles of excess reactant = initial moles - moles used in the reaction
moles of excess reactant = 0.172 mol - (2 * moles of NO4 produced)
moles of excess reactant = 0.172 mol - (2 * 0.256 mol)
moles of excess reactant ≈ -0.34 mol

Since the result is negative, it means that all of the excess reactant (O2) is consumed in the reaction. Therefore, there will be 0 liters of excess reactant left.

Answer 3

To determine the amount of excess reactant left after the reaction, we need to first find the limiting reactant. The limiting reactant is the reactant that is completely consumed first and determines the amount of product formed.

Let's start by writing the balanced chemical equation for the reaction:

2 NO(g) + O2(g) -> 2 NO2(g)

According to the stoichiometry of the balanced equation, 2 moles of nitrogen monoxide (NO) react with 1 mole of oxygen (O2) to produce 2 moles of nitrogen dioxide (NO2).

Next, let's calculate the number of moles of each reactant using the ideal gas law equation:

PV = nRT

Rearranging the equation, we have:

n = PV / RT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

Given:
Volume of nitrogen monoxide (NO) = 3.00 liters
Volume of oxygen (O2) = 2.00 liters
Pressure (P) = 695 torr
Temperature (T) = 27 degrees Celsius

First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 27 + 273.15 = 300.15 K

Now, let's calculate the number of moles of each reactant:

For nitrogen monoxide (NO):
P = 695 torr = 695/760 atm (converting torr to atm)
V = 3.00 liters
T = 300.15 K

n(NO) = (695/760) * (3.00) / (0.0821 * 300.15)

For oxygen (O2):
P = 695 torr = 695/760 atm (converting torr to atm)
V = 2.00 liters
T = 300.15 K

n(O2) = (695/760) * (2.00) / (0.0821 * 300.15)

Now, compare the mole ratios between NO and O2. The reactant with the smaller number of moles is the limiting reactant.

After determining the limiting reactant, we can calculate the moles of nitrogen dioxide (NO2) formed using the stoichiometric ratio. From there, we can find the volume of the excess reactant using the ideal gas law equation. The difference between the given volume and the volume of NO or O2 used is the amount of excess reactant remaining.

By following these steps, you can find the answer to your question.