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Comprehension # 8 `A` factory, producing methanol, is based on the reaction: `CO+2H_(2)hArrCH_(3)OH` Hydrogen `&` carbon monoxide are obtained by the reaction `CH_(4)+H_(2)OhArrCO+3H_(2)` Three units of factory namely, the "reformer" for the `H_(2)` and `CO` production, the "methanol reactor" for production of methanol and a "separator" to separate `CH_(3)OH` from `CO` and `H_(2)` are schematically shown in figure. four positions are indicated as `alpha,beta,gamma` and `delta`. The flow of methanol at position `gamma` is `10^(3)mol//sec`. The factory is so designed that `(2)/(3)` of the `CO` is converted to `CH_(3)OH`. Excess of `CO` and `H_(2)` at position `delta` are used to heat the first reaction. Assume that the reformer reaction goes to completion. At the position `(beta)` mole ratio of `CO` to `H_(2)` is `(1)/(3)` `CO+2H_(2)hArrCH_(3)OH " "DeltaH_(r)=-100 R` What is the flow of `CO` and `H_(2)` at position `(beta)` ?A. `CO : 1500mol//sec., H_(2) : 2000 mol//sec`.B. `CO : 1500mol//sec., H_(2) : 3000 mol//sec`.C. `CO : 1000mol//sec., H_(2) : 2000 mol//sec`.D. `CO : 1500mol//sec., H_(2) : 4500 mol//sec`. |
Answer» Correct Answer - D `{:(,CH_(4)+H_(2)OrarrCO,+,3H_(2)),(," "(3)/(2)xx10^(3),,3):}` `CO+2H_(2) rarr CH_(3)OH` `10^(3)" "2xx10^(3)" "10^(3)("moles/sec")` `("total" CO = (3)/(2) xx 10^(3))` |
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