Question

Using the ideal equation of state A reaction between liquid reactants takes place at 2.0 °C in a sealed, evacuated vessel with a measured volume of 15.0 L Measurements show that the reaction produced 34. g of boron trifluoride gas the pressure of boron trifluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits. pressure: at

Answer 1

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Temperature,

Volume, V = 15 L

Mass of Boron triflouride produced is 34 gm.

Molar mass of boron triflouride = 11+19*3 = 11+57 = 68 gm/mol

Hence, number of moles of BF3 , n = 34 gm/ 68 gm/mol = 0.5 mol

Assuming the gas obeys ideal gas laws,

$pV = nRT \\ p = \frac{nRT}{V}$

Where p = pressure, in atm

R = gas constant =

Putting all the values together,

$p = \frac{0.5 \ mol \times .082 \ L \ atm \ K^{-1} \ mol^{-1} \times 275 \ K}{15 \ L}=0.75 \ atm$

hence, the pressure of BF3 gas after the reaction is 0.75 atm.