What volume of 0.250 M HNO3 (nitric acid) reacts with 44.8 mL of a 0.150 M Na₂CO3 (sodium carbonate) solution? 2 HNO3(aq) + Na₂CO3 (aq) - 2 NaNO3(aq) + H₂O (1) + CO₂ (g)

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Chapter1: Chemical Foundations
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**Problem Statement:**

What volume of 0.250 M HNO₃ (nitric acid) reacts with 44.8 mL of a 0.150 M Na₂CO₃ (sodium carbonate) solution?

**Chemical Equation:**

\[ 2 \, \text{HNO}_3 \, (\text{aq}) + \text{Na}_2\text{CO}_3 \, (\text{aq}) \rightarrow 2 \, \text{NaNO}_3 \, (\text{aq}) + \text{H}_2\text{O} \, (\text{l}) + \text{CO}_2 \, (\text{g}) \]

**Explanation:**

This equation represents the chemical reaction between nitric acid (HNO₃) and sodium carbonate (Na₂CO₃). In this reaction, two moles of aqueous nitric acid react with one mole of aqueous sodium carbonate to produce two moles of aqueous sodium nitrate (NaNO₃), one mole of liquid water (H₂O), and one mole of gaseous carbon dioxide (CO₂). 

To solve for the volume of HNO₃ required, use stoichiometry based on the given concentrations and volume of Na₂CO₃.
Transcribed Image Text:**Problem Statement:** What volume of 0.250 M HNO₃ (nitric acid) reacts with 44.8 mL of a 0.150 M Na₂CO₃ (sodium carbonate) solution? **Chemical Equation:** \[ 2 \, \text{HNO}_3 \, (\text{aq}) + \text{Na}_2\text{CO}_3 \, (\text{aq}) \rightarrow 2 \, \text{NaNO}_3 \, (\text{aq}) + \text{H}_2\text{O} \, (\text{l}) + \text{CO}_2 \, (\text{g}) \] **Explanation:** This equation represents the chemical reaction between nitric acid (HNO₃) and sodium carbonate (Na₂CO₃). In this reaction, two moles of aqueous nitric acid react with one mole of aqueous sodium carbonate to produce two moles of aqueous sodium nitrate (NaNO₃), one mole of liquid water (H₂O), and one mole of gaseous carbon dioxide (CO₂). To solve for the volume of HNO₃ required, use stoichiometry based on the given concentrations and volume of Na₂CO₃.
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