2.00 g of strontium bicarbonate [Sr(HCO3)2] reacts with 250 mL of 6.00 M HCl to produce carbon dioxide, water, and strontium chloride. Assuming the reaction proceeds to completion and the volume of solution remains constant: What is the final concentration of chloride ions in solution? What is the final concentration of strontium ions in solution? Note: Strontium chloride is soluble in water. Sr(HCO3)2 + 2 HCl —–> 2 CO2 + 2 H2O + SrCl2
The Correct Answer and Explanation is:
To solve this problem, we will follow these steps:
Step 1: Write the balanced chemical equation
Sr(HCO3)2+2HCl→2CO2+2H2O+SrCl2\text{Sr(HCO}_3\text{)}_2 + 2\text{HCl} \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + \text{SrCl}_2Sr(HCO3)2+2HCl→2CO2+2H2O+SrCl2
From the equation, 1 mole of strontium bicarbonate reacts with 2 moles of HCl to form 1 mole of SrCl₂.
Step 2: Determine the moles of each reactant
Molar mass of Sr(HCO₃)₂:
- Sr = 87.62 g/mol
- H = 1.01 × 2 = 2.02 g/mol
- C = 12.01 × 2 = 24.02 g/mol
- O₃ = 16.00 × 6 = 96.00 g/mol
- Total = 87.62 + 2.02 + 24.02 + 96.00 = 209.66 g/mol
Moles of Sr(HCO₃)₂:2.00 g209.66 g/mol≈0.00954 mol\frac{2.00\text{ g}}{209.66\text{ g/mol}} \approx 0.00954\text{ mol}209.66 g/mol2.00 g≈0.00954 mol
Moles of HCl:Molarity×Volume=6.00 mol/L×0.250 L=1.50 mol\text{Molarity} \times \text{Volume} = 6.00\text{ mol/L} \times 0.250\text{ L} = 1.50\text{ mol}Molarity×Volume=6.00 mol/L×0.250 L=1.50 mol
Step 3: Determine the limiting reagent
From the balanced equation, 1 mole of Sr(HCO₃)₂ requires 2 moles of HCl.
- Required HCl = 0.00954 mol × 2 = 0.01908 mol
- Available HCl = 1.50 mol
- So HCl is in excess, and Sr(HCO₃)₂ is the limiting reagent
Step 4: Calculate final concentrations
Moles of SrCl₂ produced = moles of Sr(HCO₃)₂ = 0.00954 mol
- SrCl₂ dissociates in water: SrCl2→Sr2++2Cl−\text{SrCl}_2 \rightarrow \text{Sr}^{2+} + 2\text{Cl}^-SrCl2→Sr2++2Cl−
So:
- Moles of Sr²⁺ = 0.00954 mol
- Moles of Cl⁻ = 2 × 0.00954 mol = 0.01908 mol from SrCl₂
- Excess HCl (unreacted): 1.50 mol−0.01908 mol=1.48092 mol1.50\text{ mol} – 0.01908\text{ mol} = 1.48092\text{ mol}1.50 mol−0.01908 mol=1.48092 mol
Total Cl⁻ ions = from SrCl₂ + from unreacted HCl
= 0.01908 mol + 1.48092 mol = 1.50 mol
Final concentrations (volume = 0.250 L):
- [Cl⁻] = 1.50 mol0.250 L=6.00 M\frac{1.50\text{ mol}}{0.250\text{ L}} = \boxed{6.00\text{ M}}0.250 L1.50 mol=6.00 M
- [Sr²⁺] = 0.00954 mol0.250 L=0.0382 M\frac{0.00954\text{ mol}}{0.250\text{ L}} = \boxed{0.0382\text{ M}}0.250 L0.00954 mol=0.0382 M
Final Answers:
- Final concentration of chloride ions: 6.00 M
- Final concentration of strontium ions: 0.0382 M
Explanation
In this chemical reaction, strontium bicarbonate reacts with hydrochloric acid to form carbon dioxide, water, and strontium chloride. The stoichiometry of the reaction indicates a 1:2 molar ratio between strontium bicarbonate and HCl. That means one mole of Sr(HCO₃)₂ needs two moles of HCl to completely react and yield one mole of SrCl₂.
We begin by calculating the moles of Sr(HCO₃)₂ using its molar mass, which is approximately 209.66 g/mol. A 2.00 g sample gives us about 0.00954 mol. Then, we calculate the moles of HCl, using the given concentration and volume, yielding 1.50 mol. Comparing these quantities shows that HCl is in excess, and strontium bicarbonate is the limiting reagent.
Next, we use the stoichiometric relationship to find how much SrCl₂ is produced. Since 0.00954 mol of Sr(HCO₃)₂ reacts fully, it produces the same amount of SrCl₂. SrCl₂ dissociates into Sr²⁺ and 2 Cl⁻ ions in solution. Additionally, the leftover HCl also contributes to the Cl⁻ concentration, since each HCl molecule provides one Cl⁻ ion.
Finally, dividing the total moles of each ion by the volume of solution (0.250 L) gives the final concentrations. Chloride ions come from both the SrCl₂ and the excess HCl, summing up to 1.50 mol, giving a final concentration of 6.00 M. The strontium ion concentration comes solely from the dissolved SrCl₂ and is 0.0382 M.
This problem demonstrates stoichiometry, limiting reagents, and solution concentration concepts.
