A propane stove burned 470. grams of propane and produced 625 milliliters of
water. What is the % yield?
C₃H₈+5O₂ -> 3CO₂+ 4H₂O
To find the percent yield, we need to compare the actual yield of water with the theoretical yield.
Step 1: Convert grams of propane to moles of propane.
To do this, we need to find the molar mass of propane (C₃H₈).
C = 12.01 g/mol
H = 1.01 g/mol
Add the molar masses together:
3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol
Now, use the molar mass to convert grams to moles:
470 g C₃H₈ x (1 mol C₃H₈ / 44.11 g C₃H₈) = 10.65 mol C₃H₈
Step 2: Use the balanced equation to calculate the theoretical yield of water.
From the balanced equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Every mole of propane (C₃H₈) produces 4 moles of water (H₂O). Therefore,
10.65 mol C₃H₈ x (4 mol H₂O / 1 mol C₃H₈) = 42.60 mol H₂O
Step 3: Convert milliliters of water to moles of water.
To do this, we need to know the density of water. The density of water is approximately 1 g/mL.
So, 625 mL of water is equal to 625 g of water.
Now, convert grams to moles:
625 g H₂O x (1 mol H₂O / 18.02 g H₂O) = 34.68 mol H₂O
Step 4: Calculate the percent yield.
Use the formula:
% yield = (actual yield / theoretical yield) x 100
% yield = (34.68 mol H₂O / 42.60 mol H₂O) x 100 = 81.45%
Therefore, the percent yield is approximately 81.45%.
To calculate the percent yield, you need to compare the actual yield to the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained based on the balanced chemical equation. In this case, the balanced chemical equation is:
C₃H₈ + 5O₂ -> 3CO₂ + 4H₂O
To calculate the theoretical yield of water, you need to convert the mass of propane burned to moles and then use the stoichiometric coefficients from the balanced equation to determine the moles of water that can be produced.
1. Convert the mass of propane to moles:
To do this, you will need to know the molar mass of propane (C₃H₈). In one mole of propane, there are 3 moles of carbon (C) and 8 moles of hydrogen (H). The molar mass of carbon is approximately 12.01 g/mol, and the molar mass of hydrogen is approximately 1.01 g/mol.
The molar mass of propane (C₃H₈) is:
3(12.01 g/mol for carbon) + 8(1.01 g/mol for hydrogen) = 44.11 g/mol
Now, divide the mass of propane burned (470 g) by the molar mass of propane to convert grams to moles:
470 g / 44.11 g/mol = 10.65 mol
2. Use the stoichiometric coefficients from the balanced equation to determine the theoretical yield of water:
From the balanced equation, you can see that 1 mole of propane (C₃H₈) produces 4 moles of water (H₂O). Therefore, the theoretical yield of water can be calculated using the mole ratio:
10.65 mol propane x (4 mol water / 1 mol propane) = 42.6 mol water
3. Convert the theoretical yield of water from moles to milliliters:
To do this conversion, you need to know the density of water. The density of water at room temperature is approximately 1 g/mL.
The mass of water produced can be calculated using the molar mass of water (H₂O), which is approximately 18.01 g/mol:
42.6 mol water x (18.01 g/mol) = 767.1 g water
Finally, convert the mass of water produced to milliliters by dividing by the density of water:
767.1 g water / 1 g/mL = 767.1 mL water
4. Calculate the percent yield:
The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100:
Percent yield = (actual yield / theoretical yield) x 100
Actual yield = 625 mL (given in the question)
Percent yield = (625 mL / 767.1 mL) x 100 = 81.4%
Therefore, the percent yield of water is 81.4%.
mols C3H8 = 470/molar mass C3H8.
Using the coefficients in the balanced equation, convert mols 3H8 to mols H2O
Now convert mols H2O to grams H2O. g = mols x molar mass. This is the theoretical yield (TY) in grams. The actual yield (AY) is 625 mL so we must convert that to grams. I assume you are expected to use 625 mL x (1 mol/22,400) = about 0.028 mols and g = mols x molar mass = 0.028 x 18 = about 0.5g.
Then %yield = (AY/TY)*100 = ?