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Mass of methanol in the mixture = 200 kg x 0.15 = 30 kg metanol Moles of methanol; b) The mixture contains 85% by weight of methyl acetate.
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Mass of the slurry = 565 g – 65 g = 500 g Volumetric flow rate of the suspension = 455 ml /min Mass flow rate of the suspension = 500 g/min Density of the suspension = 500 g / 455 ml = 1.099 g/ml Mass of the calcium carbonate = 215 g – 65 g = 150 g Mass of the water in the suspension = 500 – 150 g = 350 g Mass fraction of the calcium carbonate in the suspension = 150 g/500 g = 0.
A mixture of methanol and methyl acetate contains 15.0 wt% methanol. a) Determine the kg-moles of methanol in 200 kg of the mixture. b) The flow rate of the methyl acetate in the mixture is to be 100 lbm/h. What must be the mixture flow rate in lbm/h?
b) The mixture contains 85% by weight of methyl acetate. 200 kg of the mixture contains 30 kg of methanol, 170 kg of methyl acetate. The molecular weight of methyl acetate is 74 kg / kmol. Molar amount of methyl acetate; 𝑛 =
Molar flow rate of methyl acetate (is already given in the question). 100
Mass flow rate of the methyl acetate 𝑚̇ = 45.
Mass flow rate the mixture; 𝑚̇ = 3357
1 𝑘𝑔 𝑘𝑎𝑟ışı𝑚
2.2046 𝑙𝑏𝑚 𝑘𝑎𝑟ışı𝑚 1 𝑘𝑔 𝑘𝑎𝑟ışı𝑚