Sodium hydroxide reacts with carbon dioxide as follows: which is the limiting reactant when 1.85 mol naoh and 1.00 mol co2 are allowed to react? how many moles of na2co3 can be produced? how many moles of the excess reactant remain after the completion of the reaction?

Question

Ashanti Barker

Chemistry

17 January, 19:52

Sodium hydroxide reacts with carbon dioxide as follows: which is the limiting reactant when 1.85 mol naoh and 1.00 mol co2 are allowed to react? how many moles of na2co3 can be produced? how many moles of the excess reactant remain after the completion of the reaction?

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Anderson · Answer 1

First, let's illustrate the problem into chemical formulas in a complete balanced reaction.

2NaOH + CO2 ⇒ Na2CO3 + H2O

To abide with the Law of Conservation of Mass, we balance the equation. We use these stoichiometric coefficients to determine the yield and which is the limiting reactant. Let's take 1.85 mol of NaOH. How many moles CO2 does it take to completely react NaOH?

1.85 mol NaOH * (1 mol CO2 / 2 mol NaOH) = 0.925 mol CO2

Since you have 1.00 mol CO2 present, it is still enough to completely react 1.85 mol NaOH. Therefore, CO2 is the excess reactant and NaOH is the limiting reactant. We base our calculations on the limiting reactant from here.

Next, how many moles of Na2CO3 are produced?

1.85 mol NaOH * (1 mol Na2CO3/2 mol NaOH) = 0.925 mol Na2CO3

Lastly, the amount of the excess reactant would be 1.00 mol CO2 - 0.925 mol CO2 (needed to react 1.85 mol NaOH) = 0.075 mol CO2 excess