How many grams of O2 are required to burn 13.0g of acetylene?
To determine how many grams of O2 are required to burn 13.0g of acetylene (C2H2), we need to write a balanced chemical equation for the combustion reaction.
The balanced equation for the combustion of acetylene is:
2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O
From the balanced equation, we can see that for every 2 moles of acetylene (C2H2), we need 5 moles of oxygen (O2) to fully complete the combustion.
1. Convert the mass of acetylene (C2H2) to moles:
- The molar mass of acetylene (C2H2) = 26.04 g/mol
- Moles = Mass / Molar Mass = 13.0 g / 26.04 g/mol = 0.499 moles (rounded to three decimal places).
2. Determine the moles of oxygen (O2) required using the mole ratio:
- According to the balanced equation, the mole ratio of acetylene (C2H2) to oxygen (O2) is 2:5.
- Moles of O2 = Moles of C2H2 * (5/2) = 0.499 moles * (5/2) = 1.247 moles (rounded to three decimal places).
3. Convert moles of oxygen (O2) to grams:
- The molar mass of oxygen (O2) = 32.00 g/mol.
- Mass = Moles * Molar Mass = 1.247 moles * 32.00 g/mol = 39.9 g (rounded to one decimal place).
Therefore, approximately 39.9 grams of O2 are required to burn 13.0 grams of acetylene (C2H2).
The chemical formula for acetylene is C2H2.
To determine the amount of oxygen (O2) required to burn 13.0g of acetylene (C2H2), we need to balance the combustion reaction between acetylene and oxygen.
The balanced equation for the combustion of acetylene is:
C2H2 + 5/2 O2 -> 2 CO2 + H2O
From the equation, we can see that 1 mole of acetylene (C2H2) requires 5/2 moles of oxygen (O2) to burn completely.
Now, let's calculate the number of moles of acetylene in 13.0g:
Molar mass of C2H2 = (12.01 g/mol * 2) + (1.01 g/mol * 2) = 26.04 g/mol
Moles of C2H2 = 13.0g / 26.04 g/mol = 0.499 moles
Since the ratio is 1:5/2, we multiply the moles of C2H2 by 5/2 to find the moles of O2 required:
Moles of O2 = 0.499 moles * (5/2) = 1.2495 moles
Finally, to convert moles to grams, we multiply the moles by the molar mass of O2:
Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol
Grams of O2 = 1.2495 moles * 32.00 g/mol = 39.98 g
Therefore, approximately 39.98 grams of O2 are required to burn 13.0 grams of acetylene.
Here is an example.
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