In the reaction 2H2 + 0, > 2H2O, if you start with 6 moles of H, and 4 moles of 02; which is the limiting reactant?
2Hz+02→2420
2. For the reaction N2H4 + 2Cl > N/Cl+ 2HCl, if you start with 2 moles of N2H4 and 3 moles of Cl, what is the theoretical yield of HCI (in moles)?
3. In the reaction Zn + 2HCI - ZnCl + H2, if you start with 10 grams of zinc (Zn), what is
the theoretical yield of zinc chloride (ZnCI) in grams? (Zn = 65.4 g/mol, Cl = 35.5
g/mol)
Answer (1)
[tex]\begin{gathered}\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}\end{gathered}[/tex]1. Limiting Reactant in the Reaction 2H₂ + O₂ → 2H₂O Given: Starting moles of H₂: 6 moles Starting moles of O₂: 4 moles Reaction Stoichiometry: From the balanced equation: 2 moles of H₂ react with 1 mole of O₂. Determine the Required Moles: To react with 6 moles of H₂: Required moles of O₂: [tex]\text{Required O}_2 = \frac{6 \text{ moles H}_2}{2} = 3 \text{ moles O}_2[/tex] Comparison: Available O₂: 4 moles Required O₂: 3 moles Since we have enough O₂ to react with H₂, the limiting reactant is: Limiting Reactant: H₂ 2. Theoretical Yield of HCl in the Reaction N₂H₄ + 2Cl₂ → N₂Cl₂ + 2HCl Given: Starting moles of N₂H₄: 2 moles Starting moles of Cl₂: 3 moles Reaction Stoichiometry: From the balanced equation: 1 mole of N₂H₄ reacts with 2 moles of Cl₂. Determine the Required Moles: To react with 2 moles of N₂H₄: Required moles of Cl₂: [tex]\text{Required Cl}_2 = 2 \times 2 = 4 \text{ moles}[/tex] Comparison: Available Cl₂: 3 moles Required Cl₂: 4 moles Since we don't have enough Cl₂, the limiting reactant is: Limiting Reactant: Cl₂ Calculate Theoretical Yield of HCl: From the balanced equation, 2 moles of Cl₂ produce 2 moles of HCl. Thus, 3 moles of Cl₂ will produce 3 moles of HCl. Theoretical Yield of HCl: 3 moles 3. Theoretical Yield of ZnCl₂ in the Reaction Zn + 2HCl → ZnCl₂ + H₂ Given: Mass of Zn: 10 grams Molar mass of Zn: 65.4 g/mol Molar mass of Cl: 35.5 g/mol (for 2 Cl in ZnCl₂, we will use 2 × 35.5) Calculate Moles of Zn: [tex]\text{Moles of Zn} = \frac{10 \text{ g}}{65.4 \text{ g/mol}} \approx 0.153 \text{ moles}[/tex] Reaction Stoichiometry: From the balanced equation: 1 mole of Zn produces 1 mole of ZnCl₂. Thus, 0.153 moles of Zn will produce 0.153 moles of ZnCl₂. Calculate Mass of ZnCl₂: Molar mass of ZnCl₂: [tex]\text{Molar Mass of ZnCl}_2 = 65.4 \text{ g/mol} + (2 \times 35.5 \text{ g/mol}) = 136.4 \text{ g/mol}[/tex] Theoretical Yield in Grams: [tex]\text{Mass of ZnCl}_2 = 0.153 \text{ moles} \times 136.4 \text{ g/mol} \approx 20.86 \text{ g}[/tex] Theoretical Yield of ZnCl₂: 20.86 grams[tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\begin{gathered} \boxed{\begin{array}{cc} \sf \footnotesize \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ···\: ʚ\: \: \: \: \: \: Hope\:it\:helps \: \: \: \: \: ɞ \:··· \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\footnotesize\:\# CarryOnLearning \\ \sf\footnotesize ૮₍´˶ \: • \: . \: • \: ⑅ ₎ა \: \leadsto \footnotesize\sf\color{purple} \underline{Study\:Well!}\end{array}}\end{gathered}[/tex][tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\large\qquad\qquad\qquad\tt MARCH/14/2025 [/tex]
