How many litres of oxygen are required to react with 23 g of methane according to the following equation ?
Ch4+2o2=co2+2h2O
The balanced equation tells us that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2) to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O).
First, we need to find the number of moles of methane present in 23 g.
Molar mass of CH4 = 12.01 + 4(1.01) = 16.05 g/mol
Number of moles of CH4 = mass/molar mass = 23/16.05 = 1.43 mol
According to the equation, 1 mole of CH4 reacts with 2 moles of O2.
Therefore, the number of moles of O2 required = 2 x 1.43 = 2.86 mol
Now, we can use the ideal gas law to find the volume of oxygen required at standard temperature and pressure (STP).
1 mole of any gas at STP occupies 22.4 L.
Therefore, 2.86 mol of oxygen will occupy 2.86 x 22.4 = 64 L.
So, 64 litres of oxygen are required to react with 23 g of methane.