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 = 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
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