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Ronch [10]
3 years ago
11

What is the molarity of 0.945 L of a solution that contains 0.227 moles of Fe(NO3)3?

Chemistry
1 answer:
Wewaii [24]3 years ago
4 0

Answer:

0.2402 M

Explanation:

0.945 L sol contains = 0.227 moles

1 L sol will contain = (0.227/0.945)*1 = 0.2402 moles

Molarity = 0.2402 moles/L or M

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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
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Answer:

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

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Let's determine the limiting reactant:

Mass / molar mass = moles

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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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