Final answer:
To burn 9340 L of acetylene, we need 3.34 tanks of oxygen, which means we require 4 tanks of oxygen to ensure complete combustion, as we round up to account for the impossibility of a partial tank.
Step-by-step explanation:
Calculating the Number of Oxygen Tanks Needed for Burning Acetylene
To determine how many tanks of oxygen will be required to burn the acetylene from an oxyacetylene welding torch, we need to consider the stoichiometry of the combustion reaction. The balanced chemical equation for the combustion of acetylene in oxygen is:
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
According to this equation, 2 moles of acetylene require 5 moles of oxygen to burn completely. Since the question provides the volumes of the gases at standard temperature and pressure (0 °C and 1 atm), where 1 mole of any gas occupies 22.4 L, we can directly compare volumes as if they were moles.
Given that one tank of acetylene provides 9340 liters of C₂H₂, we can set up a ratio to find the volume of O₂ needed:
(9340 L C₂H₂) / (2 L C₂H₂) × (5 L O₂) = 23,350 L O₂
Now we divide the total volume of oxygen needed by the volume one tank of oxygen provides, which is 7.00 × 10³ L:
(23,350 L O₂) / (7.00 × 10³ L O₂/tank) = 3.34 tanks
Since we cannot have a fraction of a tank, we would need 4 tanks of oxygen to ensure that there's enough oxygen to burn all the acetylene.