C2H5OH (alcohol) + Al2O3 catalyst goes to C2H4 (ethene) + H2O
when ethanol is converted to ethene a 60% yield can be expected. Assuming this yield, what is the maximum number of 75cm^3 test tubes of ethene that could be collected at room teperature and pressure when 2.4cm^3 of ethanol, density 0.8cm^-3 , react?
could someone please help me understand how to do this question?
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C2H5OH → C2H4 + H2O
Supposing the missing unit on the density of ethanol to be “grams”:
(2.4 cm^3 C2H5OH) x (0.8 g·cm^-3) / (46.06867 g C2H5OH/mol) x
(1 mol C2H4 / 1 mol C2H5OH) x (0.60) = 0.02501 mol C2H4
Supposing room temperature to be 25°C:
(0.02501 mol) x (22.414 L/mol) x (25 + 273) / (0 + 273) = 0.6119 L at 25°C
(611.9 mL) / (75 ml/test tube) = 8 (and a small fraction) test tubes