16 gram of CaC2 and 18 gram of H2O were mixed in the given reaction.
CaC2 + 2H2O it yields C2H2 + Ca(OH)2.
First, you need to determine the number of moles of each reactant:
1. Calculate the molecular weights of the reactants:
- CaC2: Ca = 40.08 g/mol, C = 12.01 g/mol
Molecular weight of CaC2 = 40.08 + (2 * 12.01) = 64.10 g/mol
- H2O: H = 1.008 g/mol, O = 16.00 g/mol
Molecular weight of H2O = (2 * 1.008) + 16.00 = 18.02 g/mol
2. Determine moles of reactants:
- CaC2: moles = mass / molecular weight = (16 g) / (64.10 g/mol) = 0.25 moles
- H2O: moles = mass / molecular weight = (18 g) / (18.02 g/mol) = 1.0 moles
3. Find the limiting reactant:
- The molar ratio of CaC2 to H2O in the balanced chemical reaction is 1:2.
- For 0.25 moles of CaC2, we would need 0.50 moles of H2O, but we have 1.0 moles available, so CaC2 is the limiting reactant.
4. Determine the moles of products formed:
- According to the balanced chemical equation, 1 mole of CaC2 produces 1 mole of C2H2 and 1 mole of Ca(OH)2.
- Since we have 0.25 moles of CaC2, we will form 0.25 moles of C2H2 and 0.25 moles of Ca(OH)2.
5. Determine the mass of products formed:
- C2H2: C = 12.01 g/mol, H = 1.008 g/mol
Molecular weight of C2H2 = (2 * 12.01) + (2 * 1.008) = 26.04 g/mol
Mass of C2H2 = moles * molecular weight = (0.25 moles) * (26.04 g/mol) = 6.51 g
- Ca(OH)2: Ca = 40.08 g/mol, O = 16.00 g/mol, H = 1.008 g/mol
Molecular weight of Ca(OH)2 = 40.08 + (2 * 16.00) + (2 * 1.008) = 74.10 g/mol
Mass of Ca(OH)2 = moles * molecular weight = (0.25 moles) * (74.10 g/mol) = 18.53 g
Thus, the given reaction will yield 6.51 g of C2H2 and 18.53 g of Ca(OH)2.