What is the % yield of of a reaction in which 40.00g of tungsten(VI)oxide (WO3) reacts with excess hydrogen gas to produce metalic tungsten and 5.40mL of water. (assume the density of water is 1.00g/mL)

Question

What is the % yield of of a reaction in which 40.00g of tungsten(VI)oxide (WO3) reacts with excess hydrogen gas to produce metalic tungsten and 5.40mL of water. (assume the density of water is 1.00g/mL)

Answer 1

WO3 + 3H2 ==> W + 3H2O

mols WO3 = grams/molar mass = approx 0.17
Using the coefficients in the balanced equation, convert mols WO3 to mols H2O. That's ?mols WO3 x (3 mols H2O/1 mol WO3) = ?mols WO3 x 3 = xx mols H2O.
Then convert xx mols H2O to grams. g = mols x molar mass. I obtained approximately 9 g. This is the theoretical (100%) yield.
% yield if 5.4 g is obtained is
% yield = (5.4/approx 9)*100 =

Answer 2

To calculate the percent yield of a reaction, we need to compare the actual amount of product obtained to the theoretical amount of product that could have been obtained.

First, let's determine the molar mass of WO3 (tungsten(VI)oxide) and H2O (water):

Molar mass of WO3:
W (tungsten) = 183.84 g/mol
O (oxygen) = 16.00 g/mol (3 atoms in WO3)
Total molar mass of WO3 = 183.84 g/mol + (16.00 g/mol x 3) = 231.84 g/mol

Molar mass of H2O:
H (hydrogen) = 1.01 g/mol (2 atoms in H2O)
O (oxygen) = 16.00 g/mol
Total molar mass of H2O = (1.01 g/mol x 2) + 16.00 g/mol = 18.02 g/mol

Next, we need to calculate the theoretical yield of tungsten (W) produced:

1 mol of WO3 produces 1 mol of W.
From the balanced chemical equation, we can see that the stoichiometric ratio is 1:1.

Convert the given mass of WO3 to moles:
40.00 g WO3 x (1 mol WO3 / 231.84 g WO3) = 0.1726 mol WO3

Since the stoichiometric ratio is 1:1, the theoretical yield of W will also be 0.1726 mol.

Now, let's calculate the actual yield of W. We are given the amount of water produced, and we can use its density to convert the volume to mass:

Given volume of water = 5.40 mL
Mass of water = 5.40 mL x 1.00 g/mL = 5.40 g

From the balanced chemical equation, we can see that for every 1 mol of W produced, 6 moles of water are produced.

Convert the mass of water to moles:
5.40 g H2O x (1 mol H2O / 18.02 g H2O) = 0.2997 mol H2O

Since the stoichiometric ratio is 1:6, the actual yield of W can be calculated as follows:

0.2997 mol H2O x (1 mol W / 6 mol H2O) = 0.0499 mol W

Finally, we can calculate the percent yield:

Percent yield = (actual yield / theoretical yield) x 100
Percent yield = (0.0499 mol W / 0.1726 mol W) x 100 = 28.9%

Therefore, the percent yield of the reaction is 28.9%.