What volume of oxygen at s.t.p is required to burn completely 7.5 dm\(^3\) of methane according to the following equation? CH\(_4\) \(_g\) + 20\(_2\) \(_g\)...

Answer Details

To find the volume of oxygen required to burn 7.5 dm\(^3\) of methane, we need to balance the chemical equation for the reaction. The balanced equation is: CH\(_4\) \(_g\) + 2O\(_2\) \(_g\) → CO\(_2\) \(_g\) + 2H\(_2\)O\(_g\) This equation tells us that for every molecule of methane that reacts, 2 molecules of oxygen are required. So, for 7.5 dm\(^3\) of methane, we need 2 * 7.5 = 15 dm\(^3\) of oxygen. So, the volume of oxygen required to burn 7.5 dm\(^3\) of methane at s.t.p is 15.0 dm\(^3\).