Final answer:
To find the mass of NO₂ formed, we need to determine the limiting reactant. Comparing the moles of reactants to the stoichiometric ratio and using the molar masses, the moles of NO and O₂ are found to be 1.53 mol and 0.84 mol, respectively. Since the ratio of NO to O₂ is 1:1, O₂ is the limiting reactant. Using the stoichiometric ratio from the balanced equation, the mass of NO₂ formed is found to be 38.64 grams.
Step-by-step explanation:
In order to find the mass of NO2 formed, we need to determine the limiting reactant first. To do this, we compare the moles of reactants to the stoichiometric ratio from the balanced equation. The molar mass of NO is 30 g/mol and the molar mass of O2 is 32 g/mol.
First, convert the given masses to moles using the molar mass:
45.8 g NO ÷ 30 g/mol = 1.53 mol NO
26.9 g O2 ÷ 32 g/mol = 0.84 mol O2
Next, compare the moles of NO and O2 to the stoichiometric ratio:
From the balanced equation, 1 mole of NO corresponds to 1 mole of O2. Therefore, the ratio of NO to O2 is 1:1.
Since the ratio of NO to O2 is 1:1, we can see that O2 is the limiting reactant because it has fewer moles than NO.
To determine the mass of NO2 formed, we use the stoichiometric ratio from the balanced equation:
1 mole of O2 reacts to form 2 moles of NO2 (from the balanced equation).
Therefore, the molar mass of NO2 is:
2 × (14 + 16 + 16) = 46 g/mol
Now, calculate the mass of NO2 formed
0.84 mol O2 × (46 g/mol) = 38.64 g NO2
So, the mass of NO2 that can be formed is 38.64 grams.