Calculate how much acetylene (C2H2) can be produced from 358 g of H20 and an excess of CaC2 if the perfect yield for this reaction is 94.5%

CaC2+H2)-> C2H2+C2H2+CA(OH)2

To calculate the amount of acetylene (C2H2) that can be produced from 358 g of H2O and an excess of CaC2 with a perfect yield of 94.5%, you need to follow these steps:

1. Determine the molar masses:
- H2O (water) has a molar mass of approximately 18.015 g/mol.
- CaC2 (calcium carbide) has a molar mass of approximately 64.099 g/mol.
- C2H2 (acetylene) has a molar mass of approximately 26.038 g/mol.

2. Convert the mass of H2O to moles:
- Divide the mass of H2O by its molar mass:
moles of H2O = 358 g / 18.015 g/mol

3. Use the stoichiometry of the balanced equation to determine the moles of C2H2:
- The balanced equation is: CaC2 + 2H2O → C2H2 + Ca(OH)2
- According to the equation, 1 mole of CaC2 produces 1 mole of C2H2.
- Since you have water in excess, the moles of C2H2 produced will be equal to the moles of CaC2 used.

4. Calculate the moles of CaC2:
- Multiply the moles of H2O by the stoichiometric ratio of CaC2 to H2O from the balanced equation. In this case, the ratio is 1:2 (1 mole of CaC2 reacts with 2 moles of H2O):
moles of CaC2 = moles of H2O * (1 mole CaC2 / 2 moles H2O)

5. Determine the perfect yield of acetylene:
- Multiply the moles of CaC2 by the perfect yield percentage (94.5% or 0.945):
perfect yield moles of C2H2 = moles of CaC2 * perfect yield

6. Convert the moles of C2H2 to grams:
- Multiply the moles of C2H2 by its molar mass:
grams of C2H2 = moles of C2H2 * molar mass of C2H2

7. The result will be the amount of acetylene (C2H2) produced in grams.

It is essential to note that the perfect yield given is theoretical and might not be the exact yield achieved in practical experiments.

This is not a limiting reagent problem. I call these simple stoichiometry problems.

Balance the equation.
Convert g H2O to mols.
Convert mols H2O to mols C2H2.
Convert mols C2H2 to grams. g = mols x molar mass.