\r\nA mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO according to the equation, CuO + H2 Ò Cu + H20 ; Oxygen then oxidizes the copper formed : Cu + ½O2 Ò CuO. 100 cm³ of the mixture measured at 25°C and 750 mm yields 84.5 cm³of dry oxygen measured at 25°C and 750 mm after passing over CuO and drying agent. What is the mole present of H2 in the mixture?
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A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO according to the equation, CuO + H2 Ò Cu + H20 ; Oxygen then oxidizes the copper formed : Cu + ½O2 Ò CuO. 100 cm³ of the mixture measured at 25°C and 750 mm yields 84.5 cm³of dry oxygen measured at 25°C and 750 mm after passing over CuO and drying agent. What is the mole present of H2 in the mixture?