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?

Explanation

`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`