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3 years ago

The density of a sample of F2(g) is 1.559 g/L. If the gas is held at 755 torr, calculate the temperature.

Chemistry

1 answer:

4vir4ik [10]3 years ago

7 0

Answer:

The temperature is 590,10 K

Explanation:

As stated, we understand that there is one mole of F2, which is equivalent to 37.9968x2 grams = 75.936 grams. The unit of Torr is converted to atm, 1 Torr = 0.0013 atm, and with the data of grams it is converted into volume through density:

density = m / v ---> v = m / density = 75,936 g / 1,559 g / l = 48, 708 l

Exactly the formula PV = nRT

n = number of moles (according to what is interpreted in the sentence is 1 mole)

P = pressure 755 Torr (converted to atm 755Torr / 0.0013 atm / Torr = 0.99344 atm)

R = value constant 0.082 l x atm / K x mol

T = is our mystery (in degrees Kelvin)

0.99344 x 48, 708 l = 1 mol x 0.082 l x atm / K x mol x X

X = 0.99344 x 48, 708 l / 1 mol x 0.082 l x atm / K x mol

X = 590, 10 K

The temperature is 590.10 K

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Given mass of aluminium = 25.0 g

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By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

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As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = $\frac{3}{2}\times 0.93=1.395moles$ of iron metal

Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

$1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g$

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