What mass of an NO2 is formed when and NO reacts with 384 g of O2?
Well, let's crunch some numbers here. The balanced equation for the reaction is:
2NO + O2 -> 2NO2
So, we can see that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. Now, to find the mass of NO2 formed, we need to find the moles of O2 first.
The molar mass of O2 is approximately 32 g/mol. So, 384 g of O2 is equal to (384 g) / (32 g/mol) = 12 moles of O2.
Since the ratio of moles of O2 to NO2 is 1:2, we know that 12 moles of O2 will produce 2 * 12 = 24 moles of NO2.
Now, the molar mass of NO2 is approximately 46 g/mol. So, the mass of NO2 formed is equal to (24 moles) * (46 g/mol) = 1104 g.
Therefore, when NO reacts with 384 g of O2, it forms 1104 g of NO2. And that's a pretty NOice amount if you ask me!
To determine the mass of NO2 formed when NO reacts with O2, we need to balance the chemical equation first. The balanced equation for the reaction is:
2NO + O2 -> 2NO2
According to the balanced equation, 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2.
Now, we need to calculate the number of moles of O2 in 384 g using its molar mass. The molar mass of O2 is 32 g/mol. We can use the formula:
moles = mass / molar mass
moles of O2 = 384 g / 32 g/mol = 12 mol
Next, we determine the theoretical yield of NO2. Since the reaction is balanced as 2 moles of NO:2 moles of NO2, the number of moles of NO2 produced will be the same as the number of moles of O2 used.
The moles of NO2 = 12 mol
Finally, we calculate the mass of NO2 using its molar mass. The molar mass of NO2 is 46 g/mol.
mass = moles × molar mass
mass of NO2 = 12 mol × 46 g/mol
mass of NO2 = 552 g
Therefore, the mass of NO2 formed when NO reacts with 384 g of O2 is 552 grams.
To determine the mass of NO2 formed when NO reacts with O2, we need to balance the chemical equation first. The balanced chemical equation for the reaction between NO and O2 is:
2 NO + O2 → 2 NO2
From the equation, we can see that for every 2 moles of NO reacted, we will obtain 2 moles of NO2 formed. Now, let's calculate the number of moles of NO2 formed step by step:
Step 1: Convert the given mass of O2 to moles.
Using the molar mass of O2 (32 g/mol), we can calculate the number of moles of O2:
384 g of O2 × (1 mol/32 g) = 12 moles of O2
Step 2: Determine the limiting reagent.
To determine the limiting reagent, we need to compare the number of moles of O2 with the reaction stoichiometry. In this case, 2 moles of NO react with 1 mole of O2. Therefore, we multiply the moles of O2 by the stoichiometric ratio:
12 moles of O2 × (2 moles of NO/1 mole of O2) = 24 moles of NO
Since we have an excess of O2 (12 moles of O2), the limiting reagent is NO.
Step 3: Calculate the number of moles of NO2 formed.
Since the stoichiometry ratio of NO:NO2 is 2:2, we find that for every 2 moles of NO reacted, we obtain 2 moles of NO2:
24 moles of NO × (2 moles of NO2/2 moles of NO) = 24 moles of NO2
Step 4: Convert moles of NO2 to mass.
Finally, we convert the moles of NO2 to mass using the molar mass of NO2 (46 g/mol):
24 moles of NO2 × (46 g/1 mol) = 1104 g of NO2
Therefore, the mass of NO2 formed when NO reacts with 384 g of O2 is 1104 grams.
How many grams of NO are required to react with this amount of O2
2NO+O2>>2NO2
the moles of NO2 will be twice the moles of O2
MolesNO2=384/32
massNO2=molesNO2*(14+32)
All this assumes NO is in excess in the reaction