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Calculate the weight of `K_2Cr_2O_7` required to produce from excess oxalic acid `(H_2C_2O_4),8.2L` `CO_2` at `127^2` and 1.0 atm pressure? |
Answer» `underset((1 mol))(Cr_2O_7^(2-))+8H^(o+)+3H_2C_2O_4underset((6 mol))(6CO_2)+2Cr^(3+)+7H_2O` `n=(PV)/(RT)=(1 atmxx 8.2L)/(0.082L" atm mol"^(-1)xx400K)` `=0.25molCO_2` `Mw(K_2Cr_2O_7)=2xx39+2xx52+16xx7=294g` 6 mol `CO_2` is obtained from 1 " mol of "`K_2Cr_2O_7` 0.25 mol `CO_2` is obtained `=(1xx0.25)/(6)=0.04` mol `=0.04xx294g=11.76g` |
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