3) What is the theoretical yield of hydrogen gas when 12.5g of sodium metal reac

Question

3) What is the theoretical yield of hydrogen gas when 12.5g of sodium metal reacts with an excess of water, while also producing sodium hydroxide? (need balanced equation). What is the limiting reactant? What is the percent yield if 0.444g of hydrogen gas are actually produced?

Explanation / Answer

2Na + 2H2O --> 2NaOH + H2 Since H2O is in excess Na is limiting reagent 2 moles of Na gives 1 mole of H2 so 12.5/23 moles will give 12.5/23*2 moles of H2 = .272 moles of H2 = .543 g of H2 Percentage yield if .444 g is obtained is .444/.543 * 100 = 81.76 %

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.