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Vikentia [17]
2 years ago
11

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

Chemistry
1 answer:
Anarel [89]2 years ago
3 0

Answer:

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

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Explanation:

When an electron is present in ground level that is the lower energy level and moves to the higher energy level, the energy is provided to the electron which means that it absorbs light. This is known as absorption.

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3 years ago
Using molecular orbital theory, explain why the removal of one electron in O2 strengthens bonding, while the removal of one elec
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