4HN3 + 5 O2 = 4NO + 6 H20.
How many grams of O2 are required to produce 90 grams of NO?
Step 1.
Convert 90 g NO to mols. #mol = g/molar mass.
Step 2.
Convert mols NO to mol O2 using the coefficients in the balanced equation.
Step 3. Convert mols O2 to grams O2. grams = # mols O2 x molar mass O2.
Post your work if you get stuck.
To determine the number of grams of O2 required to produce 90 grams of NO, we can use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.
The balanced chemical equation is:
4 HN3 + 5 O2 → 4 NO + 6 H2O
From the equation, we can see that 4 moles of HN3 react with 5 moles of O2 to produce 4 moles of NO. Thus, the molar ratio between O2 and NO is 5:4.
To calculate the grams of O2 required, we need to convert the given mass of NO (90 grams) into moles using the molar mass of NO.
The molar mass of NO is:
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of NO = (14.01 g/mol) + (16.00 g/mol) = 30.01 g/mol
Now, we can calculate the number of moles of NO:
moles of NO = mass of NO / molar mass of NO
moles of NO = 90 g / 30.01 g/mol ≈ 2.999 mol
Since the molar ratio between O2 and NO is 5:4, we can set up the following proportion:
moles of O2 / moles of NO = 5 / 4
Solving for moles of O2:
moles of O2 = (moles of NO × 5) / 4
moles of O2 = (2.999 mol × 5) / 4 ≈ 3.748 mol
Finally, to calculate the mass of O2, we can use the molar mass of O2:
molar mass of O2 = 32.00 g/mol
mass of O2 = moles of O2 × molar mass of O2
mass of O2 = 3.748 mol × 32.00 g/mol ≈ 119.936 g
Therefore, approximately 119.936 grams of O2 are required to produce 90 grams of NO.
To determine the number of grams of O2 required to produce 90 grams of NO, we first need to calculate the molar masses of NO and O2.
The molar mass of NO can be calculated by adding the atomic masses of nitrogen (N) and oxygen (O).
The atomic mass of N is 14.01 g/mol, and the atomic mass of O is 16.00 g/mol.
Therefore, the molar mass of NO is 14.01 g/mol + 16.00 g/mol = 30.01 g/mol.
Similarly, the molar mass of O2 is the sum of the atomic masses of two oxygen atoms, which can be calculated as:
2 x 16.00 g/mol = 32.00 g/mol.
Next, we need to set up a molar ratio relationship between the reactant (O2) and the product (NO) based on the balanced chemical equation.
From the balanced equation, we can see that:
4 moles of O2 react to produce 4 moles of NO.
Using the molar mass of NO, we can convert the grams of NO to moles:
90 g NO x (1 mol NO / 30.01 g NO) = 2.999 mol NO (rounded to 3 mol NO)
Now, we can use the molar ratio to determine the number of moles of O2 required to produce 3 moles of NO.
Since the balanced equation shows that 4 moles of O2 react to produce 4 moles of NO, we can set up a proportion:
4 moles O2 / 4 moles NO = X moles O2 / 3 moles NO
Cross-multiplying, we get:
4 moles O2 x 3 moles NO = 4 moles NO x X moles O2
12 moles O2 = 4 moles NO x X moles O2
Solving for X, we find:
X = 12 moles O2 / 4 moles NO = 3 moles O2
Finally, we can convert the moles of O2 to grams by multiplying by the molar mass of O2:
3 moles O2 x (32.00 g O2 / 1 mol O2) = 96.00 g O2
Therefore, 90 grams of NO would require approximately 96 grams of O2.