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Chapter 3: Problem 75

Sodium hydroxide reacts with carbon dioxide as follows: \(2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) Which is the limiting reactant when \(1.85 \mathrm{~mol} \mathrm{NaOH}\) and 1.00 \(\mathrm{mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

Short Answer

Expert verified
The limiting reactant is NaOH. A total of 0.925 moles of Na2CO3 can be produced, and 0.075 moles of CO2 remain after the completion of the reaction.

Step by step solution

01

Understand the Balanced Chemical Equation

The given balanced chemical equation is: \[2 NaOH (s) + CO_{2} (g) \longrightarrow Na_{2} CO_{3} (s) + H_{2}O (l)\] The coefficients in the balanced equation give the mole ratios between reactants and products. In this case: - 2 moles of NaOH react with 1 mole of CO2 to produce 1 mole of Na2CO3 and 1 mole of H2O.
02

Analyze the Given Quantities of Reactants

We are given the following quantities of reactants: - 1.85 moles of NaOH - 1.00 moles of CO2
03

Determine the Limiting Reactant and the Excess Reactant

We need to compare the given quantities of reactants with the stoichiometric ratios from the balanced equation. Calculate the amount of NaOH needed to react completely with 1.00 moles of CO2: \(1.00 \, moles \, CO_{2} \times \frac{2 \, moles \, NaOH}{1 \, mole \, CO_{2}} = 2.00 \, moles \, NaOH\) However, we only have 1.85 moles of NaOH available. Therefore, the limiting reactant will be NaOH, as there is not enough NaOH to completely react with the given amount of CO2. The excess reactant will be CO2.
04

Calculate the Number of Moles of Na2CO3 Produced

As NaOH is the limiting reactant, the number of moles of Na2CO3 produced can be found using its stoichiometric ratio with NaOH: \(1.85 \, moles \, NaOH \times \frac{1 \, mole \, Na_{2}CO_{3}}{2 \, moles \, NaOH} = 0.925 \, moles \, Na_{2}CO_{3}\) So, 0.925 moles of Na2CO3 can be produced.
05

Calculate the Number of Moles of Excess Reactant Remaining

To find the remaining amount of CO2, we need to see how much of it reacted with the limiting reactant, NaOH: \(1.85 \, moles \, NaOH \times \frac{1 \, mole \, CO_{2}}{2 \, moles \, NaOH} = 0.925 \, moles \, CO_{2} \, reacted\) Now subtract the reacted amount of CO2 from the initial amount: \(1.00 \, mole \, CO_{2} - 0.925 \, moles \, CO_{2} = 0.075\, moles \, CO_{2}\, remaining\) Therefore, 0.075 moles of CO2 remain after the reaction is complete. In summary, the limiting reactant is NaOH, 0.925 moles of Na2CO3 can be produced, and 0.075 moles of CO2 remain after the completion of the reaction.

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Most popular questions from this chapter

Section 2.9 introduced the idea of structural isomerism, with 1-propanol and 2 -propanol as examples. Determine which of these properties would distinguish these two substances: (a) boiling point; (b) combustion analysis results; (c) molecular weight; (d) density at a given temperature and pressure. You can check on the properties of these two compounds in Wolfram Alpha (http://www.wolframalpha.com/) or the CRC Handbook of Chemistry and Physics.

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