2h2 (g) + o2 (g) → 2h2o (l) what is the theoretical yield of water?  

• Theoretical yield of water = 5.63 g
• mass = moles x molar mass
• Limiting reagent => O₂ = 0.15625 moles
• Excess reagent = H₂ by 2.48 moles - 0.3125 moles = 2.1675 moles
• Limiting reagent dictates the amount of product formed.

Step-by-step explanation

• Balanced reaction is as:
• 2 H₂ (g) + O₂ (g) ==> 2 H₂O (l)
• We are given
• mass H₂ = 5.0 g
• mass O₂ = 5.0 g
• Part A => Calculating moles of reactants
• we know that respective molar masses H₂ = 2.016 g/mole and O₂ = 32 g/mole
• mass = moles x molar mass thus
• moles = mass / molar mass thus
• moles H₂ = 5.0 g / 2.016 g/mole
• moles H₂ = 2.48 moles
• moles O₂ = 5.0 g/ 32 g/mole
• moles O₂ = 0.15625 moles
• Part B => Finding the limiting and excess reagents
• we have calculated
• moles H₂ = 2.48 moles
• moles O₂ = 0.15625 moles
• The limiting reagent gets depleted first thus tops the reaction from continuing, in doing this it dictates the amount of product formed
• From the balanced reaction we can see that:
• 2 H₂ (g) + O₂ (g) ==> 2 H₂O (l)
• Mole ratio H₂:O₂ = 2:1
• thus
• moles H₂ needed to react with 0.15625 moles O₂ = 2/1 x 0.15625 moles = 0.3125 moles ( we have 2.48 moles H₂)
• moles O₂ needed to completely react with2.48 moles H₂ = 1/2 x 2.48 moles = 1.24 moles (we have 0.15625 moles O₂)
• We can see that, O₂ gets depleted first thus
• Limiting reagent => O₂
• Excess reagent = H₂ by 2.48 - 0.3125 moles = 2.1675 moles
• Part C => Finding mass of water formed / theoretical yield
• We know
• Limiting reagent => O₂ = 0.15625 moles
• Thus it dictates the amount of water formed
• from the balanced reaction,
• 2 H₂ (g) + O₂ (g) ==> 2 H₂O (l)
• mole ratio O₂ : H₂O = 1:2 thus
• moles H₂O produced = 2/1 x 0.15625 moles
• moles H₂O = 0.3125 moles
• We know
• mass = moles x molar mass (18.015 g/mole)
• mass H₂O = 0.3125moles x 18.015 g/mole
• mass H₂O = 5.63 g
• This is the theoretical yield of water