Grams of butane (C4H10) formed by the liquefaction of 448 litres of the gas (measured at (STP) would be
A. 580
B. 640
C. 1160
D. Data insufficient; can't be computed
Answer: Option C
Solution(By Examveda Team)
At S.T.P 1 mole of any gas occupies $$22.4\,\,ltrs$$Given $$448\,\,ltrs$$ so, number of moles = $$\frac{{448}}{{22.4}}$$ $$= 20$$
We know number of moles = $$\frac{{{\text{weight}}}}{{{\text{gram moleculear weight}}}}$$ $$\left( {{\text{molecular weight of }}{C_4}{H_{10}} = 58} \right)$$
So, $$20 =$$ $$\frac{{{\text{weight}}}}{{58}}$$
Hence weight of $${C_4}{H_{10}} = 1160$$
This Question Belongs to Chemical Engineering >> Chemical Engineering Thermodynamics
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