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When 1.7 mol of NaOH and 1.0 mol of CO₂ are allowed to react, how many moles of Na₂CO₃ can be produced?

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Final answer:

Using the balanced chemical equation, 2 NaOH + CO₂ → Na₂CO₃ + H₂O, it can be determined that 1.0 mole of CO₂ will react with 1.7 moles of NaOH in excess to produce 1.0 mole of Na₂CO₃.

Step-by-step explanation:

The student is asking about the stoichiometry of a chemical reaction involving sodium hydroxide (NaOH) and carbon dioxide (CO₂) to produce sodium carbonate (Na₂CO₃). To determine how many moles of Na₂CO₃ can be produced, the balanced chemical equation is needed:

2 NaOH(aq) + CO₂(aq) → Na₂CO₃(aq) + H₂O(l)

From the balanced equation, it is clear that 2 moles of NaOH react with 1 mole of CO₂ to produce 1 mole of Na₂CO₃. The student has 1.7 moles of NaOH and 1.0 mole of CO₂. Since the NaOH is in excess (1.7 > 1 * 2), the limiting reactant is CO₂, and it will determine the amount of Na₂CO₃ produced.

Therefore, 1.0 mole of CO₂ will produce 1.0 mole of Na₂CO₃, as per the stoichiometry of the reaction.

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