To solve this problem, we can use the stoichiometry of the reaction between acetylene (C2H2) and oxygen (O2) to determine the volume of O2 needed.
The balanced chemical equation for the combustion of acetylene is:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
This equation shows that 2 moles of acetylene react with 5 moles of oxygen to produce 4 moles of carbon dioxide and 2 moles of water.
From the equation, we can see that the ratio of C2H2 to O2 is 2:5. Therefore, if the volume of acetylene is 40 L, we can calculate the volume of oxygen needed as follows:
V(O2) = (40 L * 5) / 2 = 100 L
Therefore, the volume of oxygen needed for the combustion of 40 L of acetylene is 100 L.
To solve this problem, we can use the stoichiometry of the reaction between acetylene (C2H2) and oxygen (O2) to determine the volume of O2 needed.
The balanced chemical equation for the combustion of acetylene is:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
This equation shows that 2 moles of acetylene react with 5 moles of oxygen to produce 4 moles of carbon dioxide and 2 moles of water.
From the equation, we can see that the ratio of C2H2 to O2 is 2:5. Therefore, if the volume of acetylene is 40 L, we can calculate the volume of oxygen needed as follows:
V(O2) = (40 L * 5) / 2 = 100 L
Therefore, the volume of oxygen needed for the combustion of 40 L of acetylene is 100 L.