Answer:
Mass of n-butane used up = 29.0 - 23.2 = 5.8 kg No. of moles of n-butane used up\[=\frac{5.8\times 1000g}{58\,g\,mo{{l}^{-1}}}\,=100\] To calculate volume of 100 moles of the gas at \[27{}^\circ C\] and 1 atm pressure, apply\[PV=nRT\] or \[V=\frac{nRT}{P}=\frac{100\,mol\,\times 0.0821\,L\,atm\,{{K}^{-1}}\,mo{{l}^{-1}}\times 300K}{1\,atm}\] \[=2463\,L=2\cdot 463\,{{m}^{3}}\,(1\,{{m}^{3}}={{10}^{3}}L)\]
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