In the reaction
2H2+O2-2H2O.
80cm^3 of hydrogen was reacted with 25cm^3 of oxygen, calculate the volume of residual gases
Before we can calculate the volume of residual gases, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To determine the limiting reactant, we compare the moles of hydrogen and oxygen present in the given volumes. Since the coefficients in the balanced equation show a 2:1 ratio between H2 and O2, we need to convert the volumes to moles before comparing.
The molar volume of any gas at standard temperature and pressure (STP) conditions is 22.4 L/mol or 22,400 cm^3/mol. We can use this conversion factor to convert the volumes to moles.
80 cm^3 of H2 x (1 L/1000 cm^3) x (1 mol/22.4 L) = 0.00357 mol of H2
25 cm^3 of O2 x (1 L/1000 cm^3) x (1 mol/22.4 L) = 0.00112 mol of O2
Since the ratio of H2 to O2 is 2:1, we have an excess of hydrogen. Therefore, oxygen is the limiting reactant.
To determine the volume of residual gases, we need to calculate the moles of oxygen consumed in the reaction:
0.00112 mol of O2 x (2 mol of H2O/1 mol of O2) x (22.4 L/mol) x (1000 cm^3/1 L) = 50.1 cm^3 of O2 consumed
Since the reaction is 2H2 + O2 -> 2H2O, the number of moles of O2 consumed is equal to the number of moles of water formed. Therefore, the volume of residual gases is equal to the initial volume of hydrogen minus the volume of water formed:
80 cm^3 of H2 - 50.1 cm^3 of H2O = 29.9 cm^3 of residual gases
So the volume of residual gases is 29.9 cm^3.