How many grams of O2(g) are needed to completely burn 34.1 g of C3H8(g)?
25.4 grams
https://www.webqc.org/balance.php?reaction=C3H8+%2B+O2+%3D+CO+%2B+H2O
To solve this question, we need to balance the chemical equation for the combustion of propane (C3H8) and then use stoichiometry to determine the amount of oxygen (O2) required.
Step 1: Balanced Chemical Equation:
The balanced equation for the combustion of propane is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 2: Calculate the molar mass:
- Molar mass of C3H8 = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol
- Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol
Step 3: Convert grams to moles:
- Moles of C3H8 = 34.1 g / 44.11 g/mol = 0.7735 mol
Step 4: Use stoichiometry:
From the balanced equation, we know that 1 mole of C3H8 requires 5 moles of O2. Therefore,
- Moles of O2 needed = 0.7735 mol C3H8 × (5 mol O2 / 1 mol C3H8) = 3.8675 mol
Step 5: Convert moles to grams:
- Grams of O2 needed = 3.8675 mol × 32.00 g/mol = 123.60 g
Therefore, 123.60 grams of O2 are needed to completely burn 34.1 grams of C3H8.