how many L @STP if oxygen are needed to produce 45.0 grams of nitric oxide according to the following balanced chemical equation? 4NH3 + 5O2=4NO + 6H2O
mols NO = grams/molar mas = ?
Using the coefficients in the balanced equation, convert mols nO to mols O2. That ratio is 4 mols NO/5 mols O2.
Then remember that one mole of a gas at STP occupies 22.4 L.
CALCULATE THE OXIDATION NUMBER FOR THE FOLLOWING CR IN K2CR2O4 AND MN IN MNO4
CALCULATE CR IN K2CR2O4 AND MN IN MNO4
To determine how many liters of oxygen (O2) are needed to produce 45.0 grams of nitric oxide (NO) according to the balanced chemical equation provided (4NH3 + 5O2 = 4NO + 6H2O), we need to follow these steps:
Step 1: Convert the given mass of nitric oxide (NO) to moles.
To do this, we need the molar mass of NO, which is 30.01 g/mol.
Using the formula:
moles = mass / molar mass
moles of NO = 45.0 g / 30.01 g/mol = 1.499 moles
Step 2: The balanced chemical equation tells us that 5 moles of O2 are required to produce 4 moles of NO. So, we can set up a ratio using the molar coefficients:
5 O2 : 4 NO
Step 3: Calculate the moles of O2 required.
Using the ratio from Step 2, we can set up a proportion to determine the moles of O2:
5 O2 / 4 NO = x O2 / 1.499 moles of NO
Cross-multiplying:
5 O2 × 1.499 moles of NO = 4 NO × x O2
x O2 = (5 O2 × 1.499 moles of NO) / 4 NO
Step 4: Calculate the volume of O2 at STP (Standard Temperature and Pressure).
At STP, 1 mole of any gas occupies 22.4 liters of volume.
Converting moles to liters:
x O2 (in liters) = (5 O2 × 1.499 moles of NO) / 4 NO × 22.4 liters/mole
Simplifying:
x O2 = 5.623 liters of O2
Therefore, approximately 5.623 liters of oxygen (O2) are needed to produce 45.0 grams of nitric oxide (NO) according to the given balanced chemical equation at STP.