What will be the theoretical yield of tungsten (is), W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

WO₃ + 3H₂ → 3H₂O + W

Let's determine the limiting reactant:

Mass / molar mass = moles

45 g / 231.84 g/mol = 0.194 moles

1.50 g / 2 g/mol = 0.75 moles

Ratio is 1:3. 1 mol of tungsten (VI) oxide needs 3 moles of hydrogen to react.

Let's make rules of three:

1 mol of tungsten (VI) oxide needs 3 moles of H₂

Then 0.194 moles of tungsten (VI) oxide would need (0.194.3) / 1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess ... Then, the limiting is the tungsten (VI) oxide)

3 moles of H₂ need 1 mol of WO₃ to react

0.75 moles of H₂ would need (0.75. 1) / 3 = 0.25 moles

It's ok. I do not have enough WO₃.

Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.

Let's convert the moles to mass (molar mass. mol)

0.194 mol. 183.84 g/mol = 35.6 g