Chemistry: For the following reaction, 4.06 grams of carbon (graphite) are allowed to react with 14.6 grams of oxygen gas

Question

Carbon (graphite) (s) + oxygen (g) —-> carbon dioxide (g)

what is the maximum mass of carbon dioxide that can be formed?

Mass=

What is the FORMULA for the limiting reactant?

What mass of the excess reagent after the reaction is complete?

Mass=

Answer 1

C(s) + O₂(g) ==> CO₂(g)

Find limiting reactant. One way to do this is to divide moles of each reactant by its coefficient in the balanced equation and whichever value is less represents the limiting reactant.

For C(s) we have: 4.06 g C x 1 mol C / 12 g = 0.338 mols C (÷1->0.338)

For O₂(g) we have: 14.6 g O₂ x 1 mol O₂ / 32 g = 0.456 mols O₂ (÷1->0.456)

Since 0.338 is less than 0.456, C(s) is the limiting reactant

Use moles of the limiting reactant (0.338 mols) to calculate amount of CO₂ that can form

0.338 mols C x 1 mol CO₂ / mol C = 0.338 mols CO₂ x 44 g CO₂/mol = 14.9 g CO₂

To find mass of excess reagent (O₂), find out how much O₂ was used up, and subtract that from how much O₂ we started with:

O₂ used = 0.338 mols C x 1 mol O₂ / mol C = 0.338 mols O₂ used up

moles of O₂ initially present = 0.456 mols

moles O₂ left over = 0.456 - 0.338 = 0.118 mols O₂ left over

mass O₂ left over = 0.112 mols O₂ x 32 g / mol = 3.78 g O₂ left over