Imagine a beaker divided down the center by a rigid membrane that is freely permeable to water but impermeable to glucose. Side
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Answer:
The rate at which sulfur dioxide is being produced is 0.90 kg/s.
Explanation:
Volume of oxygen gas consumed in second ,V= 994 L
Pressure of the gas = p
Temperature of the gas = T = 170°C= 170 + 273 K=443 K
Moles of oxygen gas consumed in a second = n
( ideapl gas equation)
n = 21.044 mole
Moles of dioxygen gas consumed per second = 21.044 mol
(Claus process)
According to reaction, 3 moles of dioxygen gives 2 moles of sulfur dioxide gas.Then 21.044 moles of dioxygen will give;
of sulfur dioxide
Mass of 14.029 moles of sulfur dioxide gas;
14.029 mol × 64 g/mol = 897.86 g
897.86 g = 0.89786 kg ≈ 0.90 kg
Mass of sulfur dioxide produced per second = 0.90 kg
The rate at which sulfur dioxide is being produced is 0.90 kg/s.