Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.
PV = nRT ; n = PV/RT
Substituting the known values,
n = (0.930 atm)(93/1000 L) / (0.0821 L.atm/mol.K)(10 + 273.15K)
n = 3.72 x 10^-3 mols
At STP, the volume of each mol of gas is equal to 22.4 L.
volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)
volume = 0.0833 L or 83.34 mL

5 0

4 years ago

Balance this equation. if a coefficient of "1" is required, choose "blank" for that box.

blondinia [14]

Step 1: Write Imbalance Equation

CH₃CHO + O₂   → CO₂ + H₂O

Step 2: Balance Carbon Atoms:
There are 2 carbon atoms at reactant side and one at product side. So multiply CO₂ with 2 to balance them. i.e.

CH₃CHO + O₂   → 2 CO₂ + H₂O

Step 3: Balance Hydrogen Atoms:
There are 4 hydrogen atoms at reactant side and 2 Hydrogen atoms at product side. So, multiply H₂O by 2 to balance Hydrogen on both sides. i.e.

  CH₃CHO + O₂   → 2 CO₂ + 2 H₂O

Step 4: Balance Oxygen Atoms:
There are 3 Oxygen atoms at reactant side and 6 Oxygen atoms at product side. In order to balance them multiply O₂ on reactant side by 2.5 (5/2). i.e

CH₃CHO + 5/2 O₂   →   2 CO₂ +  2 H₂O

Step 6: Eliminate Fraction:
Multiply overall equation by 2 to eliminate fraction. i.e.

2 CH₃CHO + 5 O₂   →   4 CO₂ +  4 H₂O

7 0

3 years ago