11. Tungsten can also be obtained by reacting tungsten
fluoride with hydrogen. The equation for this reaction is
WF6 + 3H2 à W + 6HF
(i) In an experiment, a chemist used 59.6 g of tungsten
fluoride. What is the maximum mass of tungsten he
could obtain from 59.6 g of tungsten fluoride?
Relative formula mass of tungsten fluoride = 298 (2)
maximum mass =
g
(ii) Starting with a different mass of tungsten fluoride, he
calculates that the mass of tungsten formed should be
52.0 g. In his experiment he actually obtains 47.5 g of
tungsten. What is the percentage yield of tungsten in
this experiment? (2)
percentage yield =
%

(i) To find the maximum mass of tungsten that can be obtained from 59.6 g of tungsten fluoride, we first need to determine the mole ratio between tungsten fluoride (WF6) and tungsten (W) in the balanced chemical equation:

1 mole of WF6 produces 1 mole of W

Next, we need to calculate the number of moles of tungsten fluoride (WF6) present in 59.6 g using its molar mass:

Molar mass of WF6 = 298 g/mol

Number of moles of WF6 = Mass / Molar mass = 59.6 g / 298 g/mol = 0.2 mol

Since the mole ratio between WF6 and W is 1:1, the number of moles of tungsten (W) produced will also be 0.2 mol.

Now, we can calculate the maximum mass of tungsten:

Maximum mass of W = Number of moles of W × Molar mass of W

= 0.2 mol × 183.84 g/mol

= 36.77 g

So, the maximum mass of tungsten that can be obtained from 59.6 g of tungsten fluoride is 36.77 g.

(ii) The percentage yield of tungsten in the experiment is calculated using the formula:

Percentage yield = (Actual yield / Theoretical yield) × 100%

Given that the actual yield is 47.5 g and the theoretical yield (calculated in part i) is 36.77 g, we can substitute these values into the formula:

Percentage yield = (47.5 g / 36.77 g) × 100%

≈ 129.22%

So, the percentage yield of tungsten in this experiment is approximately 129.22%.