Answer to Question #312912 in General Chemistry for Ibrahim

Question

If 25.0cm3 of methane was burned in 20cm3 of oxygen

I. Write an equation to represent the reaction

II. Which of the gas was in excess

III. Calculate the volume of the excess gas

Answer 1

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methan - 2 mole oxygen, so

22,4 l - 1 mole of methan

0.025 l - x mole

n(CH₄)=0.00112 mole

0.02 l - x mole

n(O₂) =0.00089 mole

So methan is in excess

0.00088 mole of oxygen can react with 0.00088/2=0.000446 mole of methan

0.00112 - 0.000446=0.000674 mole is in excess.

0.000674 mole is 0.0146 l = 15 cm³