Limiting Reactant-Excess Reactant - Sodium Hydroxide And Carbon Dioxide Reaction
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Limiting Reactant-Excess Reactant - Sodium Hydroxide And Carbon Dioxide Reaction Sodium hydroxide reacts with carbon dioxide as follows:

2NaOH(s) + CO2(g) ----> Na2CO3(s) + H2O(l).

(a) Define the terms limiting reactant and excess reactant
(b) Why are the amounts of products formed in a reaction determined only by the amount of limiting reactant.
(c) Which reagent is the limiting reactant when 1.85 moles of NaOH reacts with 1 mole of CO2?
(d) How many moles of Na2CO3 can be produced?
(e) How many moles of the excess reactant remain after the completion of the reaction?

The strings: S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).

The math:
Although we are calculating , mass, grams and moles, the chemical reaction is the reason for the calculation. So Pj Problem of Interest is of type interaction (chemical). (a) The amounts of products that can be produced in a chemical reaction, depend on the availability of the reactants in the reaction. Sometimes, the mole-ratio of reactants in a chemical reaction is less that the mole-ratio indicated in the balanced chemical equation of the reaction. The reactant that does not meet its mole-ratio requirement is the limiting reactant. The excess reactant is the amount of reactant left over after the limiting reactant has been used up.

(b) Once the limiting reactant is used up, the reaction stops. Consequently, no more products can be produced.

(c) The NaOH is the limiting reactant.

(d) 2 moles of NaOH produces 1 mole of Na2CO3
So, 1.85 moles of NaOH will produce 1.85/2 = 0.925 mol of Na2CO3

(e) 2 moles of NaOH reacts with 1 mole CO2
So, 1.85 moles of NaOH will react with 1.85/2 = 0.925 mol of CO2
Excess reactant = 1 - 0.925 = 0.075 mol of CO2.

Math

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