Given the following reaction: c3h8+5o2=3co2+4h20 how many grams of co2 will be produced 7 g of c3h8 and 98 g of o2

Answer : 21.12 g

explanation :

- A limiting reactant is the substance that is totally consumed when the chemical reaction is complete and the reaction cannot continue without it.

- This is a limiting reactant problem because the amount of product (CO2) formed is limited by this substance (either O2 or C3H8).

- We can predict the limiting reactant by calculating number of mole for all reactants;

      n (C3H8) = mass/ molar mass = 7/[(12 × 3)+(1 ×8)] = 0.16 mol

      n (O2) = mass/ molar mass = 98/(16 × 2) = 3 mol

But we know from this balanced equation that for the reaction to continue, for (n) of propane there must be (5n) of oxygen. Clearly we have more oxygen than required for the reaction to continue as

[ 3 O2 mol >  (0.16 propane mol × 5) . Hence, the limiting reactant is propane.

- Using cross multiplication,

                0.16 mol propane    →        1 mol propane

                            ?                    →        3 mol CO2

So (n) of CO2 produced = 0.16 × 3 / 1 = 0.48 mol  ,

And mass of CO2 produced = n × molar mass = 0.48 × [12 + (16 ×2)] = 21.12 g..

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