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igor_vitrenko [27]
1 year ago
10

What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a

t stp?
Chemistry
1 answer:
Alexandra [31]1 year ago
4 0

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm  [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K  [At STP]

Now first find the number of moles

F₂  +  CaBr₂  →  CaF₂  +  Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So,  199.89 g CaBr₂ reacts with  = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

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<h3>Further explanation</h3>

Given

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