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elixir [45]
3 years ago
10

In a mixture of helium and chlorine, occupying a volume of 14.6 l at 871.7 mmhg and 28.6oc, it is found that the partial pressur

e of chlorine is 355 mmhg. what is the total mass of the sample?
Chemistry
1 answer:
Inga [223]3 years ago
3 0
First, let's find the total moles.

PV=nRT
(871.7 mmHg)(1 atm/760 mmHg)(14.6 L) = n(0.0821 L-atm/mol-K)(28.6 +273)
Solving for n,
n = 0.676 moles

Assuming ideal gas behavior, the pressure fraction is also equal to mole fraction.

Mole fraction of Chlorine = 355/871.7 = 0.407
Mole fraction of Helium = 1 - 0.407 = 0.593

Knowing Chlorine to be 35.45g/mol and Helium to be 4 g/mol.
Mass of Chlorine = (0.407)(0.676)(35.45 g/mol) = 9.75 g
Mass of Helium = (0.593)(0.676)(4 g/mol) = 1.6 g
Total Mass = 9.75+1.6 = <em>11.35 g</em>
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Explanation:

First, we have to write our equation. It's actually pretty straightforward - first we look for our reactants (looks like it's Fe₂O₃ and CO), then we look for our products (Fe and CO₂). Then, we have to balance it so that both sides have the same number of both element.

Currently, we have the equation Fe₂O₃ + CO ⇒ Fe + CO₂. There are 2 Fe atoms, 4 O atoms, and 1 C atom on the left side. There is 1 Fe atom, 2 O atoms, and 1 C atom on the right side.

First thing we can do is give our Fe on the right side a coefficient of 2. This will make it equivalent to the 2 Fe atoms on the left side:

Fe₂O₃ + CO ⇒ 2Fe + CO₂

Next, we need to make sure that we have the same number of C and O atoms on each side. This takes a little bit of thinking, but what we have to do is give CO a coefficient of 3 and CO₂ a coefficient of 3. This gives us 6 O atoms on the left side (when we include the O₃) and 6 O atoms on the right side (since there are 3 O₂ atoms and 3 times 2 is 6). Here's what that looks like:

Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂

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Answer:

Explanation:

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Answer:

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Explanation:

HOW TO CALCULATE MASS OF A SUBSTANCE:

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Mass (g) = no. of moles (mol) × molar mass (g/mol)

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