Hydrochloric acid (HCI) reacts with sodium carbonate (Na
CO
), forming sodium chloride (NaCl), water (H
O), and carbon dioxide (CO
). This equation is balanced as written: 2HCl(aq) + Na
CO
(aq) ?2NaCl(aq) + H
O(l) + CO
(g) Part A What volume of 2.75 mol L
HCI in litres is needed to react completely (with nothing left over) with 0.500 L of 0.300 mol L
Na
CO
? Express your answer numerically in litres. View Available Hint(s) Submit Part B A 393-mL sample of unknown HCI solution reacts completely with Na
CO
to form 14.1 g CO
. What was the concentration of the HCl solution? Express the molar concentration numerically.
The Correct Answer and Explanation is:
Let’s break this down into two parts:
Part A
We are given the following information:
- The reaction is:
2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g)2HCl(aq) + Na_2CO_3(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g) - Volume of sodium carbonate (Na₂CO₃) = 0.500 L
- Concentration of Na₂CO₃ = 0.300 mol/L
- Concentration of HCl = 2.75 mol/L
We need to determine the volume of HCl required to react completely with Na₂CO₃.
Step 1: Calculate moles of Na₂CO₃
The number of moles of Na₂CO₃ is given by: moles of Na2CO3=concentration×volume=0.300 mol/L×0.500 L=0.150 mol\text{moles of Na}_2CO_3 = \text{concentration} \times \text{volume} = 0.300 \, \text{mol/L} \times 0.500 \, \text{L} = 0.150 \, \text{mol}moles of Na2CO3=concentration×volume=0.300mol/L×0.500L=0.150mol
Step 2: Use the stoichiometric ratio
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Na₂CO₃. This gives us the ratio: moles of HClmoles of Na2CO3=2so,moles of HCl=2×0.150 mol=0.300 mol\frac{\text{moles of HCl}}{\text{moles of Na}_2CO_3} = 2 \quad \text{so,} \quad \text{moles of HCl} = 2 \times 0.150 \, \text{mol} = 0.300 \, \text{mol}moles of Na2CO3moles of HCl=2so,moles of HCl=2×0.150mol=0.300mol
Step 3: Calculate volume of HCl
Now that we know the moles of HCl required, we can find the volume using the molarity equation: Volume of HCl=moles of HClconcentration of HCl=0.300 mol2.75 mol/L=0.109 L=109 mL\text{Volume of HCl} = \frac{\text{moles of HCl}}{\text{concentration of HCl}} = \frac{0.300 \, \text{mol}}{2.75 \, \text{mol/L}} = 0.109 \, \text{L} = 109 \, \text{mL}Volume of HCl=concentration of HClmoles of HCl=2.75mol/L0.300mol=0.109L=109mL
Part B
We are given the following information:
- Volume of unknown HCl solution = 393 mL = 0.393 L
- Mass of CO₂ produced = 14.1 g
- Molar mass of CO₂ = 44.01 g/mol
Step 1: Calculate moles of CO₂ produced
First, we calculate the moles of CO₂ produced using the given mass: moles of CO2=mass of CO2molar mass of CO2=14.1 g44.01 g/mol=0.3205 mol\text{moles of CO}_2 = \frac{\text{mass of CO}_2}{\text{molar mass of CO}_2} = \frac{14.1 \, \text{g}}{44.01 \, \text{g/mol}} = 0.3205 \, \text{mol}moles of CO2=molar mass of CO2mass of CO2=44.01g/mol14.1g=0.3205mol
Step 2: Use the stoichiometric ratio
From the balanced equation, we see that 2 moles of HCl produce 1 mole of CO₂. Therefore, the moles of HCl required are: moles of HCl=2×0.3205 mol=0.641 mol\text{moles of HCl} = 2 \times 0.3205 \, \text{mol} = 0.641 \, \text{mol}moles of HCl=2×0.3205mol=0.641mol
Step 3: Calculate the concentration of HCl
Now, we can find the molar concentration of the HCl solution: Concentration of HCl=moles of HClvolume of solution in L=0.641 mol0.393 L=1.63 mol/L\text{Concentration of HCl} = \frac{\text{moles of HCl}}{\text{volume of solution in L}} = \frac{0.641 \, \text{mol}}{0.393 \, \text{L}} = 1.63 \, \text{mol/L}Concentration of HCl=volume of solution in Lmoles of HCl=0.393L0.641mol=1.63mol/L
Final Answers:
- Part A: The volume of 2.75 mol/L HCl required is 109 mL.
- Part B: The concentration of the unknown HCl solution is 1.63 mol/L.
