What is the theoretical yield of NaCl when 23 g of Na reacts with excess Cl2?
Practice Questions
1 question
Q1
What is the theoretical yield of NaCl when 23 g of Na reacts with excess Cl2?
58.5 g
23 g
46 g
11.5 g
23 g of Na = 1 mole. 1 mole of Na produces 1 mole of NaCl = 58.5 g.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the theoretical yield of NaCl when 23 g of Na reacts with excess Cl2?
Solution: 23 g of Na = 1 mole. 1 mole of Na produces 1 mole of NaCl = 58.5 g.
Steps: 7
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 between Na and Cl2 to form NaCl: 2 Na + Cl2 → 2 NaCl.
Step 4: From the balanced equation, note 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.
