Final answer:
The amount of calcium hydroxide formed in the reaction of 4.44 g of calcium oxide with 7.77 g of water is 5.87 g. The limiting reactant is CaO, and the excess reactant is H2O, with 6.34 g of water left over after the reaction.
Step-by-step explanation:
To calculate the amount of calcium hydroxide (Ca(OH)2) formed, we need to start with the balanced equation for the reaction of calcium oxide (CaO) with water (H2O):
CaO(s) + H2O(l) → Ca(OH)2(s)
Each mole of CaO reacts with one mole of H2O to produce one mole of Ca(OH)2. To find the limiting reactant, calculate the moles of each reactant:
- Number of moles of CaO = mass (4.44 g) / molar mass (56.08 g/mol) = 0.0792 moles
- Number of moles of H2O = mass (7.77 g) / molar mass (18.02 g/mol) = 0.431 moles
Since the stoichiometry is 1:1, the limiting reactant is CaO, which will completely react, leaving some water unreacted. Now calculate the mass of Ca(OH)2 formed:
- Mass of Ca(OH)2 = moles of CaO (0.0792) × molar mass of Ca(OH)2 (74.09 g/mol) = 5.87 g
The excess reactant is water (H2O), and to find the amount left over:
- Moles of water reacted = moles of CaO = 0.0792 moles
- Total moles of water - moles reacted = 0.431 - 0.0792 = 0.3518 moles
- Mass of water left = moles left × molar mass of H2O = 0.3518 × 18.02 g/mol = 6.34 g