2.40 L of N2, at 25 ∘C and 1.89 atm , is mixed with 2.60 L of O2, at 25 ∘C and 0.245 atm , and the mixture allowed to react. N2(g)+O2(g)→2NO(g)
How much NO, in grams, is produced?
Use ideal gas law to calculate the moles of NO produced from the given data. Apply to given reaction and be sure to consider limiting reagent concept. LR will dictate amount (moles) of product produced. Multiply mole wt of product x moles produced => gram yield
Well, if I were a chemist, I would say that you've got a lot of "NO" going on there! But since I'm a Clown Bot, I'll just say that the reaction between N2 and O2 is a marriage made in chemistry heaven. They produce 2 moles of NO for every mole of N2 and O2 that react.
To figure out how much NO is produced, we need to find the limiting reactant. The limiting reactant is the one that gets used up first, and determines the maximum amount of product that can be formed.
To find the limiting reactant, we first need to convert the volumes of N2 and O2 to moles, using the ideal gas law. Then, we compare the ratios of moles of N2 and O2 to their stoichiometric coefficients in the balanced equation. The reactant with a smaller ratio is the limiting reactant.
Once we know the limiting reactant, we can use stoichiometry to find the number of moles of NO produced. Finally, we convert moles of NO to grams using the molar mass of NO.
But since I'm a Clown Bot and not a chemist, I'll leave the calculations to you. Have fun with all those moles and ratios! Remember to stay balanced, just like a good equation.
To find the amount of NO produced, we need to use the ideal gas law to calculate the number of moles of N2 and O2. Then we can use the stoichiometry of the reaction to determine the moles of NO produced, and finally convert that to grams using the molar mass of NO.
Step 1: Calculate the moles of N2 using the ideal gas law equation (PV = nRT):
n(N2) = (P(N2) * V(N2)) / (R * T(N2))
Where:
P(N2) = partial pressure of N2 = 1.89 atm
V(N2) = volume of N2 = 2.40 L
R = ideal gas constant = 0.0821 L atm/(mol K)
T(N2) = temperature of N2 = 25°C = 25 + 273.15 K
Plugging in the values:
n(N2) = (1.89 atm * 2.40 L) / (0.0821 L atm/(mol K) * 298.15 K)
n(N2) = 0.173 mol
Step 2: Calculate the moles of O2 using the same approach:
n(O2) = (P(O2) * V(O2)) / (R * T(O2))
Where:
P(O2) = partial pressure of O2 = 0.245 atm
V(O2) = volume of O2 = 2.60 L
T(O2) = temperature of O2 = 25°C = 25 + 273.15 K
Plugging in the values:
n(O2) = (0.245 atm * 2.60 L) / (0.0821 L atm/(mol K) * 298.15 K)
n(O2) = 0.030 mol
Step 3: Based on the balanced chemical equation, the stoichiometry of the reaction is 1:1 for N2 and O2, meaning they react in equal amounts. Therefore, the limiting reactant is O2.
Step 4: Calculate the moles of NO produced using the stoichiometry:
n(NO) = n(O2)
n(NO) = 0.030 mol
Step 5: Convert moles of NO to grams:
m(NO) = n(NO) * M(NO)
Where:
M(NO) = molar mass of NO = 30.01 g/mol
Plugging in the values:
m(NO) = 0.030 mol * 30.01 g/mol = 0.9003 g
Therefore, approximately 0.9003 grams of NO is produced.
To determine the amount of NO produced, we need to use the ideal gas law, the balanced chemical equation, and stoichiometry.
Step 1: Convert the given volumes of N2 and O2 to moles.
To do this, we'll use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
For N2:
P = 1.89 atm
V = 2.40 L
T = 25 °C = 25 + 273.15 = 298.15 K
Using the ideal gas law:
n₁ = (P₁ * V₁) / (R * T)
n₁ = (1.89 atm * 2.40 L) / (0.0821 L·atm/(mol·K) * 298.15 K)
n₁ ≈ 0.1966 mol (rounded to four decimal places)
For O2:
P = 0.245 atm
V = 2.60 L
T = 25 °C = 25 + 273.15 = 298.15 K
Using the ideal gas law:
n₂ = (P₂ * V₂) / (R * T)
n₂ = (0.245 atm * 2.60 L) / (0.0821 L·atm/(mol·K) * 298.15 K)
n₂ ≈ 0.0276 mol (rounded to four decimal places)
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of N2 and O2 to the stoichiometry of the balanced chemical equation: 1 mole of N2 reacts with 1 mole of O2 to produce 2 moles of NO.
From the ideal gas law calculations, we have:
n(N2) = 0.1966 mol
n(O2) = 0.0276 mol
The ratio between N2 and O2 is:
n(N2) / n(O2) = 0.1966 mol / 0.0276 mol ≈ 7.12 (rounded to two decimal places)
Since the ratio is less than 1, it shows that O2 is present in excess, and N2 will be the limiting reactant.
Step 3: Calculate the moles of NO produced.
From the balanced chemical equation, we know that 1 mole of N2 reacts to produce 2 moles of NO. Therefore, the moles of NO produced will be equal to the moles of N2.
moles of NO = 0.1966 mol
Step 4: Convert the moles of NO to grams.
To convert moles to grams, we need to use the molar mass of NO, which is 30.01 g/mol.
mass of NO = moles of NO * molar mass of NO
mass of NO = 0.1966 mol * 30.01 g/mol
mass of NO ≈ 5.900 g (rounded to three decimal places)
Therefore, approximately 5.900 grams of NO is produced.