Answer :
The maximum mass of NO2 that can be formed when 38.8 g of NO and 26.9 g of O2 react together is 59.492 grams, with the calculation based on stoichiometry and the balanced chemical equation.
Calculating the Mass of NO2 Formed in a Reaction
To calculate the mass of NO2 that can be formed given the masses of NO and O2, we'll need to perform stoichiometric calculations based on the balanced chemical equation 2 NO (g) + O2 (g)
ightarrow 2 NO2 (g). First, we'll convert the given masses to moles using the molar masses of NO (30.01 g/mol) and O2 (32.00 g/mol).
- Determine moles of NO: 38.8 g / 30.01 g/mol = 1.293 moles of NO.
- Determine moles of O2: 26.9 g / 32.00 g/mol = 0.8406 moles of O2.
- Compare the mole ratio from the balanced equation to the moles present.
- Identify the limiting reactant. In this case, NO is the limiting reactant since it would require 2 moles of NO per 1 mole of O2, and we don't have enough NO to fully react with all the O2.
- Calculate the moles of NO2 produced: Since the mole ratio of NO to NO2 is 1:1, we'll have 1.293 moles of NO2 formed.
- Calculate the mass of NO2: 1.293 moles * 46.01 g/mol (molar mass of NO2) = 59.492 g of NO2.
The maximum mass of NO2 that can be formed in this reaction is 59.492 grams.
The mass of NO2 that can be formed from the reaction of 38.8 g of NO and 26.9 g of O2 is 77.3 g, considering O2 as the limiting reactant.
To determine the mass of NO2 that can be formed when 38.8 g of NO and 26.9 g of O2 react together, we first need to calculate the molar masses of the reactants and product. The molar mass of NO is 30.01 g/mol (14.01 for nitrogen and 16.00 for oxygen) and the molar mass of O2 is 32.00 g/mol. Similarly, the molar mass of NO2 is 46.01 g/mol (14.01 for nitrogen and 2 *16.00 for oxygen).
According to the balanced chemical equation 2 NO (g) + O2 (g) → 2 NO2 (g), 2 moles of NO react with 1 mole of O2 to form 2 moles of NO2. We can convert the mass of NO and O2 to moles and find the limiting reactant, which will determine the maximum amount of NO2 that can be produced.
For NO: 38.8 g NO * (1 mol NO / 30.01 g NO) = 1.29 moles of NO
For O2: 26.9 g O2 * (1 mol O2 / 32.00 g O2) = 0.840 moles of O2
Based on the stoichiometry of the reaction, the ratio is 2 moles of NO for 1 mole of O2. Therefore, the maximum moles of NO2 that can be produced will be double the moles of O2, due to the 2:1 ratio.
Maximum moles of NO2 = 0.840 moles of O2 * 2 = 1.680 moles NO2
Mass of NO2 that can be formed = 1.680 mol * 46.01 g/mol = 77.3 g of NO2
