Question

Chapter 9 Homework ± Core Chemistry Skill: Calculating the Quantity of a Reactant or Product for a Chemical Reaction in Solution 9 of 10 Review | Constants | Periodic Table The stoichiometric relationship between the reactants and products of an aqueous reaction can be used to determine different kinds of information about a reaction, such as the volume of a given molarity reactant required to produce a certain amount of product. Hydrochloric acid (HCl) reacts with sodium carbonate (Na2CO3), forming sodium chloride (NaCl), water (H2O), and carbon dioxide (CO2). This equation is balanced as written: 2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g) Part A What volume of 1.25 M HCl in liters is needed to react completely (with nothing left over) with 0.250 L of 0.200 M Na2CO3? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) Previous Answers Incorrect; Try Again; 2 attempts remaining Part B A 435-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 11.1 g CO2. What was the concentration of the HCl solution? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) Previous Answers Incorrect; Try Again; 3 attempts remaining Provide Feedback

Answer 1

Part A

What volume of 1.25 M HCl in liters is needed to react completely (with nothing left over) with 0.250 L of 0.200 M Na2CO3? Express your answer to three significant figures and include the appropriate units.

Answer: 0.0800 L (OR) 80.0 mL

Balanced equation is: 2HCl + Na₂CO₃ ==> 2NaCl + H₂O + CO₂

moles Na2CO3 = 0.25 L x 0.2 M = 0.05 mol

moles HCl required = 2 x 0.05 = 0.1 mol

volume required = 0.1 mol / (1.25 mol / L) = 0.08 L [OR] 80 mL

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Part B

A 435-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 11.1 g CO2. What was the concentration of the HCl solution? Express your answer to three significant figures and include the appropriate units.

Answer: 1.16 M

Balanced equation is: 2HCl + Na₂CO₃ ==> 2NaCl + H₂O + CO₂

moles CO₂ = 11.1 g / 44.01 g/mol = 0.25221540559 mol

2 mol HCl produces 1 mol CO₂, so moles HCl consumed = 2 x 0.25221540559 = 0.50443081117 mol

molarity of HCl = 0.50443081117 mol / [(435 mL / 1000 mL) x 1L] = 1.15961106018 M ~ 1.16 M

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Important Note: Enter values exactly as is to avoid attempt lost, 0.0800 has three significant figures including zero's after number 8.

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