What mass of NO2 is formed when NO reacts with 384 g of 02?
Please be VERY descriptive. I've been out of school and I'm expected to take a test tomorrow on this.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To answer this question, we need to apply the concept of stoichiometry and use balanced chemical equations. In this case, we are given that NO reacts with 384 grams of O2.
Step 1: Write the balanced chemical equation:
2 NO + O2 -> 2 NO2
This equation tells us that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2.
Step 2: Convert the given mass of O2 to moles:
To do this, we need to determine the molar mass of O2 (Oxygen gas). Oxygen has an atomic mass of 16 g/mol, and since O2 consists of two oxygen atoms, the molar mass of O2 is 32 g/mol.
Now, we can calculate the number of moles of O2:
Number of moles = mass / molar mass
Number of moles = 384 g / 32 g/mol = 12 moles
Step 3: Determine the moles of NO2 produced:
Using the balanced equation, we can see that 1 mole of O2 reacts to produce 2 moles of NO2. Therefore, we can conclude that 12 moles of O2 will produce 24 moles of NO2.
Step 4: Convert moles of NO2 to mass:
To calculate the mass of NO2 produced, we need to use the molar mass of NO2. Nitrogen has an atomic mass of 14 g/mol, and Oxygen has an atomic mass of 16 g/mol. Therefore, the molar mass of NO2 is (14 + (2 * 16)) = 46 g/mol.
Now, we can calculate the mass of NO2:
Mass = number of moles * molar mass
Mass = 24 moles * 46 g/mol = 1,104 g
Therefore, when NO reacts with 384 g of O2, the mass of NO2 formed is 1,104 grams.