How many grams of carbon dioxide are produced when 16.0 g of methane and 48.0 g of oxygen gas combust molar mass of CH4 = 16.0 g/mol molar mass of O2 = 32.0 g/mol molar mass of CO2 = 44.0 g/mol?
Flexi Says:
The chemical equation for the combustion of methane is:
CH4+2O2→CO2+2H2O
Given that 16g of methane reacts with 48g of oxygen to produce carbon dioxide and water:
Number of moles of methane =
1616=1mol
Number of moles of oxygen =
4832=1.5mol
Now, the limiting reagent can be identified by dividing the number of moles of each reactant by their coefficient. So, for methane, it is 1/1 = 1 while for oxygen, it is 1.5/2 = 0.75. Thus, oxygen is the limiting reagent. Therefore, it is 0.75 moles of carbon dioxide which is produced in the chemical reaction.
Amount of carbon dioxide produced in grams =
0.75 mol×44 g/mol=33 g