Question

When 0.110 mol of carbon is burned in a closed vessel with 9.71 g of oxygen, how many grams of carbon dioxide can form?

Which reactant is in excess, and how many grams of it remain after the reaction?

Answer 1

Lets calculate the moles of oxygen first:

molar mass of oxygen = 32 g/mol

moles = mass / molar mass = 9.71 g/ 32g/mol = 0.3034 mol

now write the reaction equation:

C + O₂ → CO₂

stoichiometric conversion factor for O₂ consumed = 1 mol C / 1 mol O₂

0.3034 mol O₂ needs =   1 mol C / 1 mol O₂ x 0.3034 mol O₂

= 0.3034 mol C

but we have only 0.110 mol C and hence it is limiting reactant and O₂ is in excess

now the conversion will be based on limiting reactant only as the reaction will stop as soon as limiting reactant is over:

0.11 mol C needs 0.11 mol O₂

remaining O₂ = 0.3034 mol - 0.11 mol = 0.1934 mol

= 0.1934 mol x 32g/mol = 6.19 g will be left after reaction completion

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