The volume of excess oxygen will 2.28 L
From the equation of the reaction:
[tex]CH_4(g) + 2O_2(g) --- > CO_2(g) + 2H_2O(g)[/tex]
The mole ratio of the two reactants is 1:2.
1 mole of gas at STP = 22.4 L
Mole of 29.2 L of methane at STP = 29.2/22.4 = 1.3 moles
Mole of 63.3 L of oxygen at STP = 63.3/22.4 = 2.8 moles.
Thus, oxygen is in excess because 1.3 x 2 = 2.6 moles is needed to react with 1.3 moles methane.
Excess mole of oxygen = 2.8 - 2.6 = 0.2 moles
Volume of 0.2 moles at STP = 22.4 x 0.2 = 2.28 L
More on stoichiometric calculations can be found here: https://brainly.com/question/27287858
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