A 0.510 g sample of strontium chloride reacts with sodium carbonate to give 0.455 g of strontium carbonate. What is the percent yield of strontium carbonate in this reaction?

• A. 82.4%
• B. 78.5%
• C. 95.6%
• D. 89.2%

Answer: (D)

Verified step by step guidance

1. Determine the balanced chemical equation for the reaction between strontium chloride (SrCl₂) and sodium carbonate (Na₂CO₃) to form strontium carbonate (SrCO₃) and sodium chloride (NaCl). The balanced equation is: SrCl₂ + Na₂CO₃ → SrCO₃ + 2 NaCl.
2. Calculate the molar mass of strontium chloride (SrCl₂) and strontium carbonate (SrCO₃) using the periodic table. The molar mass of SrCl₂ is approximately 158.53 g/mol, and the molar mass of SrCO₃ is approximately 147.63 g/mol.
3. Convert the mass of the strontium chloride sample (0.510 g) to moles using its molar mass. Use the formula: moles of SrCl₂ = mass of SrCl₂ / molar mass of SrCl₂.
4. Using the stoichiometry of the balanced equation, determine the theoretical yield of strontium carbonate. Since the reaction is a 1:1 ratio, the moles of SrCO₃ produced will be equal to the moles of SrCl₂ reacted. Convert the moles of SrCO₃ to grams using its molar mass.
5. Calculate the percent yield of strontium carbonate using the formula: percent yield = (actual yield / theoretical yield) × 100%. The actual yield is given as 0.455 g.