What will be the theoretcal yield of tungsten (s), W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2? WO3 (s) + 3H2 (g) = > W (s) + 3 H2O (g).

Question

Biology

Kathleen Stephenson

6 May, 05:09

What will be the theoretcal yield of tungsten (s), W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2? WO3 (s) + 3H2 (g) = > W (s) + 3 H2O (g).

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Answers (1)

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Graves · Answer 1

Theoretical yield of tungsten produced = 35.6836915592 ≈ 35. 68 g

Explanation:

The chemical equation can be expressed as follows;

WO3 (s) + 3H2 (g) → W (s) + 3 H20 (g)

Note that the equation is already balanced.

Molecular Mass of WO3 = 183.84 + 15.999 * 3 = 183.84 + 47.997 = 231.837 g

From the equation 1 mole of WO3 reacts with 3 mole of hydrogen molecule.

Molecular mass of tungsten (W) = 183.84 g

1 mole of tungsten was produced from the chemical equation.

WO3 (s) + 3H2 (g) → W (s) + 3 H20 (g)

From the equation,

231. 837 g of WO3 produces 183.84 g of tungsten

45.0 g of WO3 will produce?

grams of tungsten produced = 183.84 * 45 / 231.837

grams of tungsten produced = 8272.8 / 231.837

Theoretical yield of tungsten produced = 35.6836915592 ≈ 35. 68 g