According to the following reaction how many moles of carbon dioxide will be formed upon the complete reaction of 28.2 grams of carbon (graphite) with excess oxygen gas?

Question

carbon (graphite) (s) + oxygen (g) #-># carbon dioxide (g)

Ellie Andrews · Accepted Answer

Each mole of carbon gives #44*g# of carbon dioxide gas upon oxidation.

#C(s) + O_2(g) rarr CO_2(g)#. The #1:1# stoichiometry is clear: moles of carbon, and dioxygen gas are equivalent to the moles of carbon dioxide gas generated. We start with #(28.2*g)/(12.01*g*mol)=2.35*mol# #C#, and thus #2.35*molxx44.01*g*mol^-1# #CO_2(g)# are evolved. Suppose that less than #100*g# of gas were collected in the experiment. Barring error on the part of the experimenter, what would be a reasonable explanation?

Ellie Andrews · Answer

undefined To determine the number of moles of carbon dioxide formed, first, we need to balance the chemical equation for the reaction between carbon (graphite) and oxygen gas:

C(graphite) + O2(g) -> CO2(g)

The balanced equation shows that 1 mole of carbon reacts with 1 mole of oxygen gas to produce 1 mole of carbon dioxide.

The molar mass of carbon is approximately 12.01 g/mol. So, 28.2 grams of carbon (graphite) is equivalent to:

28.2 g / 12.01 g/mol = 2.35 moles of carbon

Since the reaction involves 1 mole of carbon dioxide for every mole of carbon, the number of moles of carbon dioxide formed will also be 2.35 moles.