Answer:
33.33 g of methane
Step-by-step explanation:
This problem is solved by utilizing the stoichiometric quantities in the balanced equation
CH4 + H2O ======== CO + 3 H2
which tell us that 3 mol H2 gas need 1 mol of mehane gas to be produced.
Using the information that 1 Kg H2 has a milleage of 80 we can solve the problem :
1 Kg H2/ 80 miles = 1000g H2/80miles = 12.5 g H2 per mile
12.5 g H2 x 1 mol/ 2 g H2 = 6.25 mol H2
6.25 mol H2 to be produced require 6.25 mol H2 x (1 mol CH4/ 3 mol H2)
= 2.083 mol CH4
2.083 mol CH4 converted to mass is
2.083 mol CH4 x 16 g CH4/ mol = 33.33 g CH4