what is the percent yield of reaction in which 455g of tungsten (VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 9.60 cm^3 of water.(density of water = 1.00 g/cm^3) ???
WO3 + 3H2 ==> 3H2O + W
mols WO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols WO3 to mols H2O.
Now convert mols H2O to grams H2O. This is the theoretical yield (TY). The actual yield (AY) is given in the problem as 9.60 grams (since the density is 1.00 g/mL).
% yield = (AY/TY)*100 = ?
To calculate the percent yield of a reaction, we need to compare the actual yield to the theoretical yield.
First, we need to determine the moles of WO3 used in the reaction. We can use the molar mass of WO3 to convert grams to moles:
Molar mass of WO3 = (183.84 g/mol W) + (3 * 16.00 g/mol O)
= 231.84 g/mol
Moles of WO3 = mass / molar mass
= 455 g / 231.84 g/mol
= 1.963 mol WO3
From the balanced chemical equation, we can see that the stoichiometric ratio between WO3 and water is 1:3. This means that for each mole of WO3, we should obtain 3 moles of water.
Moles of water produced = moles of WO3 * (3 moles H2O / 1 mole WO3)
= 1.963 mol * 3
= 5.889 mol H2O
Next, let's calculate the volume of water produced in liters:
Volume of water produced = 9.60 cm^3 * (1 L / 1000 cm^3)
= 0.0096 L
We can now calculate the moles of water based on its volume and density. Since the density of water is 1.00 g/cm^3, the mass of the water can be determined:
Mass of water = Volume of water * Density of water
= 0.0096 L * 1.00 g/cm^3
= 0.0096 g
Moles of water produced = mass of water / molar mass of water
= 0.0096 g / 18.02 g/mol
= 0.000533 mol H2O
Finally, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100
Theoretical yield is the moles of water based on the stoichiometry of the reaction, which we have calculated to be 5.889 mol.
Percent yield = (0.000533 mol / 5.889 mol) * 100
= 0.0091 * 100
= 0.91%
Therefore, the percent yield of the reaction is approximately 0.91%.
To determine the percent yield of a reaction, you need to compare the actual yield (obtained experimentally) to the theoretical yield (calculated using stoichiometry).
Let's start by writing the balanced equation for the reaction:
WO3 + 3H2 → W + 3H2O
We can see that 1 mole of WO3 reacts with 3 moles of H2 to produce 1 mole of W and 3 moles of H2O.
1. Calculate the moles of WO3:
Given mass of WO3 = 455 g
Atomic mass of WO3 = 231.84 g/mol (tungsten) + 3 * 16.00 g/mol (oxygen) = 231.84 g/mol + 48.00 g/mol = 279.84 g/mol
Moles of WO3 = mass of WO3 / molar mass of WO3
= 455 g / 279.84 g/mol
≈ 1.626 mol
2. Calculate the moles of H2O:
Volume of water = 9.60 cm^3
Density of water = 1.00 g/cm^3
Mass of water = volume of water × density of water
= 9.60 cm^3 × 1.00 g/cm^3
= 9.60 g
Moles of H2O = mass of H2O / molar mass of H2O
= 9.60 g / 18.02 g/mol
≈ 0.532 mol
3. Determine the limiting reactant:
Since WO3 and H2 are in a 1:3 molar ratio, we compare the moles of reactants. The reactant with a smaller mole value is the limiting reactant.
According to the balanced equation, 1 mole of WO3 produces 3 moles of H2O.
So, if all the WO3 reacts, the expected yield of H2O would be 3 * 1.626 mol ≈ 4.878 mol.
Since we have obtained 0.532 mol of H2O, it is the limiting reactant, and WO3 is in excess.
4. Calculate the moles of W (theoretical yield):
According to the balanced equation, the molar ratio between WO3 and W is 1:1.
Therefore, the moles of W produced would be equal to the moles of WO3 used.
Moles of W = Moles of WO3
≈ 1.626 mol
5. Convert the moles of W to grams (theoretical yield):
Molar mass of W = 183.84 g/mol
Theoretical yield of W = moles of W × molar mass of W
= 1.626 mol × 183.84 g/mol
≈ 298.802 g
6. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100
Actual yield is the mass of W obtained from the reaction, which is not provided in the question. Once you obtain the actual yield from experiment, you can substitute it in the equation and calculate the percent yield.