What is the theoretical yield of NaCl when 23 g of Na reacts with Cl2?

Practice Questions

Q1
What is the theoretical yield of NaCl when 23 g of Na reacts with Cl2?
  1. 58.5 g
  2. 23 g
  3. 46 g
  4. 69 g

Questions & Step-by-Step Solutions

What is the theoretical yield of NaCl when 23 g of Na reacts with Cl2?
  • Step 1: Determine the molar mass of sodium (Na). The molar mass of Na is approximately 23 g/mol.
  • Step 2: Calculate the number of moles of Na in 23 g. Use the formula: moles = mass (g) / molar mass (g/mol). So, moles of Na = 23 g / 23 g/mol = 1 mole.
  • Step 3: Write the balanced chemical equation for the reaction: 2 Na + Cl2 → 2 NaCl. This shows that 2 moles of Na produce 2 moles of NaCl.
  • Step 4: From the balanced equation, we see that 1 mole of Na produces 1 mole of NaCl.
  • Step 5: Since we have 1 mole of Na, it will produce 1 mole of NaCl.
  • Step 6: Calculate the mass of NaCl produced. The molar mass of NaCl is approximately 58.5 g/mol. Therefore, 1 mole of NaCl = 58.5 g.
  • Step 7: Conclude that the theoretical yield of NaCl when 23 g of Na reacts with Cl2 is 58.5 g.
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