What is the percent yield of a reaction in which 74.1 g of tungsten(vi) oxide (wo3) reacts with excess hydrogen gas to produce metallic tungsten and 8.01 ml of water (d = 1.00 g/ml)?

Answer : The percent yield of the reaction is, 46.49 %

Explanation :  Given,

Mass of = 74.1 g

Molar mass of = 231.84 g/mole

Molar mass of = 18 g/mole

First we have to calculate the moles of .

Now we have to calculate the moles of .

The balanced chemical reaction will be,

From the balanced reaction, we conclude that

As, 1 mole of react to give 3 moles of

So, 0.319 moles of react to give moles of

Now we have to calculate the mass of

The theoretical yield of  = 17.226 g

Now we have to calculate the actual yield of water.

The actual yield of  = 8.01 g

Now we have to calculate the percent yield of

Therefore, the percent yield of the reaction is, 46.49 %

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