Calculate how much acetylene (C2H2) can be produced from 358 g of H2O and an excess of CaC2 if the percent yield for this reaction is 94.5%.

CaC2 + 2H2O -> C2H2 + Ca(OH)2

To calculate the amount of acetylene (C2H2) produced, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed and restricts the amount of product that can be formed.

1. Determine the molar mass of H2O (18 g/mol) and CaC2 (64 g/mol).

2. Convert the mass of H2O to moles:
Moles of H2O = 358 g / 18 g/mol = 19.9 mol

3. Since CaC2 is in excess, it will not limit the reaction. However, we still need to calculate the amount of it used to find the actual yield later.

4. Find the theoretical yield of C2H2. The balanced equation tells us that 1 mole of CaC2 reacts to produce 1 mole of C2H2. So, the theoretical yield of C2H2 is equal to the moles of H2O used.

Theoretical yield of C2H2 = 19.9 mol

5. Calculate the actual yield using the percent yield given (94.5%).

Actual yield = Percent yield × Theoretical yield
Actual yield = 0.945 × 19.9 mol = 18.8 mol

Therefore, 18.8 moles of acetylene (C2H2) can be produced from 358 g of H2O, assuming a percent yield of 94.5%.

To calculate the amount of acetylene (C2H2) produced, you need to follow these steps:

1. Convert the given mass of water (H2O) to moles. The molar mass of water (H2O) is 18.02 g/mol.
Moles of H2O = Mass of H2O / Molar mass of H2O

Moles of H2O = 358 g / 18.02 g/mol = 19.9 mol

2. Use the stoichiometry of the balanced equation to determine the moles of acetylene (C2H2) produced. According to the balanced equation, 1 mole of CaC2 reacts with 2 moles of water (H2O) to produce 1 mole of acetylene (C2H2).
Moles of C2H2 = 0.945 × Moles of H2O

(Since the percent yield is given as 94.5%, we multiply it by the moles of H2O to get the actual moles of C2H2 produced.)

Moles of C2H2 = 0.945 × 19.9 mol = 18.77 mol

3. Convert the moles of C2H2 to grams using the molar mass of acetylene (C2H2). The molar mass of C2H2 is 26.04 g/mol.
Mass of C2H2 = Moles of C2H2 × Molar mass of C2H2

Mass of C2H2 = 18.77 mol × 26.04 g/mol = 488.51 g

Therefore, approximately 488.51 grams of acetylene (C2H2) can be produced from 358 grams of water (H2O) with a percent yield of 94.5%.

This is just another stoichiometry problem like the HNO3/NO problem below with a slight twist.

Solve for theoretical yield in grams, as before, then
%yield = (AY/TY)*100 = 94.5 and solve for AY.