Asked by iceman
If 65.8 grams Cl2 react with excess Na, what is the percent yield if an experiment produced 99.9 g NaCl ?
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Balanced reaction: 2 Na + Cl2 → 2 NaCl
1) Moles Cl2 = 65.8 g / 70.90 g·mol⁻¹ = 0.928 mol
2) Moles NaCl theoretical = 2 × 0.928 = 1.856 mol
3) Mass NaCl theoretical = 1.856 mol × 58.44 g·mol⁻¹ = 108.5 g
Percent yield = (99.9 g / 108.5 g) × 100% = 92.1% (≈92%)
1) Moles Cl2 = 65.8 g / 70.90 g·mol⁻¹ = 0.928 mol
2) Moles NaCl theoretical = 2 × 0.928 = 1.856 mol
3) Mass NaCl theoretical = 1.856 mol × 58.44 g·mol⁻¹ = 108.5 g
Percent yield = (99.9 g / 108.5 g) × 100% = 92.1% (≈92%)
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