If 6.14 grams of nitrogen monoxide gas were reacted with 2.53 grams of oxygen, how many grams of nitrogen dioxide gas would be produced?
To solve this problem, we need to determine the molar ratio between nitrogen monoxide (NO) and nitrogen dioxide (NO2) in the balanced chemical equation. The balanced equation for the reaction between nitrogen monoxide and oxygen to produce nitrogen dioxide is:
2 NO + O2 -> 2 NO2
Here, the coefficient in front of nitrogen monoxide (NO) is 2, which means that 2 moles of NO are required to produce 2 moles of NO2.
To find the number of moles of NO and O2, we can use the formula:
moles = mass / molar mass
First, let's calculate the number of moles for each reactant:
Molar mass of NO = 14.01 g/mol for nitrogen + 16.00 g/mol for oxygen = 30.01 g/mol
moles of NO = 6.14 g / 30.01 g/mol ≈ 0.205 mol
Molar mass of O2 = 16.00 g/mol for oxygen x 2 = 32.00 g/mol
moles of O2 = 2.53 g / 32.00 g/mol ≈ 0.079 mol
Now, we need to determine which reactant is the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed. We can do this by comparing the mole ratios of the reactants.
From the balanced equation, we can see that for every 2 moles of NO, we need 1 mole of O2. Since we have 0.205 moles of NO and 0.079 moles of O2, we can see that we have more than enough O2 to react with the available NO.
Therefore, NO is the limiting reactant. This means that all of the NO will be consumed, and we will calculate the amount of product based on the moles of NO.
Since the molar ratio between NO and NO2 is 2:2, we know that for every 2 moles of NO consumed, 2 moles of NO2 are produced.
moles of NO2 = (moles of NO) x (2 moles of NO2 / 2 moles of NO)
= 0.205 mol x (2 mol NO2 / 2 mol NO)
= 0.205 mol
To calculate the mass of NO2 produced, we can use the formula:
mass = moles × molar mass
molar mass of NO2 = 14.01 g/mol for nitrogen + 16.00 g/mol for oxygen x 2 = 46.01 g/mol
mass of NO2 = 0.205 mol × 46.01 g/mol ≈ 9.43 g
Therefore, approximately 9.43 grams of nitrogen dioxide gas would be produced.
To solve this problem, we need to balance the chemical equation for the reaction between nitrogen monoxide (NO) and oxygen (O2) to produce nitrogen dioxide (NO2).
The balanced chemical equation is as follows:
2NO + O2 -> 2NO2
Based on the equation, we can see that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. Now, let's go step by step to find the answer:
Step 1: Calculate the number of moles of nitrogen monoxide (NO) and oxygen (O2) given the mass.
To find the number of moles, we will use the molar mass of each compound.
Molar mass of NO:
Nitrogen (N) = 14.01 g/mol
Oxygen (O) = 16.00 g/mol
Total molar mass of NO = 14.01 + 16.00 = 30.01 g/mol
Molar mass of O2:
Oxygen (O) = 16.00 g/mol
Total molar mass of O2 = 16.00 * 2 = 32.00 g/mol
Moles of NO = Mass of NO / Molar mass of NO
Moles of NO = 6.14 g / 30.01 g/mol ≈ 0.2046 mol
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 2.53 g / 32.00 g/mol ≈ 0.0791 mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, compare the moles of each reactant in the balanced equation. The limiting reactant is the one with a smaller number of moles.
From the balanced equation:
2 moles of NO react with 1 mole of O2.
Since the ratio of NO to O2 is 2:1, we can see that 1 mole of O2 is required for every 2 moles of NO.
Given that we have 0.2046 mol of NO and 0.0791 mol of O2, we can conclude that O2 is our limiting reactant because we have fewer moles of O2 compared to the required ratio.
Step 3: Calculate the number of moles of nitrogen dioxide (NO2) produced.
According to the balanced equation, for every 2 moles of NO, we form 2 moles of NO2.
Moles of NO2 = Moles of O2 (limiting reactant) * (2 moles of NO2 / 1 mole of O2)
Moles of NO2 = 0.0791 mol * (2 mol NO2 / 1 mol O2) = 0.1582 mol
Step 4: Convert moles of NO2 to grams of NO2.
Molar mass of NO2:
Nitrogen (N) = 14.01 g/mol
Oxygen (O) = 16.00 g/mol (two oxygen atoms)
Total molar mass of NO2 = 14.01 + (16.00 * 2) = 46.01 g/mol
Mass of NO2 = Moles of NO2 * Molar mass of NO2
Mass of NO2 = 0.1582 mol * 46.01 g/mol ≈ 7.28 g
Therefore, approximately 7.28 grams of nitrogen dioxide gas would be produced.