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iVinArrow [24]
3 years ago
14

what volume (L) of fluorine gas is required to react with 2.31 g of calcium bromide to form calcium fluoride and bromine gas at

8.19 atm and 35 degrees celsius? (Correct asnwer: 3.57x10^-3) but i got 3.57x10^-2
Chemistry
1 answer:
MrRa [10]3 years ago
5 0

Hey there!:

Balanced equation:

F2(g) + CaBr2(s) ---> CaF2 + Br2

1 mol F2 = 1 mol CaBr2

Calculation:

2.31 g CaBr2 * (1 mol CaBr2 / 199.886 g CaBr2)  * (1 mol F2 / 1 mol CaBr2)

= 0.0115565 moles of F2:

Assuming F2 is an ideal  

gas at these conditions:

P*V = n*R*T

Solving for V:

V = (n *R* T) / P

where :

n = 0.0115565 moles

R = 0.08206 atm·L/mol·K

Temperature in K = 35 + 273.15 => 308.15 K

P = 8.19 atm

Substituting numbers into V = (n x R x T) / P:

V = 0.0115565 * 0.08206 * 308.15 / 8.19

V = 0.29222 / 8.19

V = 0.0357 or 3.57*10⁻² L

You are correct!!!


Hope that helps!

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<u>____________________________</u>

Explanation:

<u>____________________________</u>

<u>Note</u>: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

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So:  We call solve using a technique known as:  "dimensional analysis" :

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  0.500 mol AlCl₃ * (\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?

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 " \frac{(0.500 * 2 * 26.98)}{2}   g Al ["grams of aluminum"] ;

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<u>Note</u>: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

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         →  Round to 3 (Three) significant figures;

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   =  13.5 g Al ; that is:  "13.5 grams of aluminum."

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 Hope this is helpful!  

      Best wishes to you in your academic pursuits—and within the "Brainly" community!

____________________________

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