1.

`Mn_(3)O_(4)` when heated with AI powered, gets reduced to produced Mn metal and `AI_(2)O_(3)`. If at least `612 g of AI_(2)O_(3) and 825g of Mn` are to be produced, the minimum amount of `Mn_(3)O_(4)` and AI required is respectively :A. `1030.5g, 324g`B. `1145g, 360g`C. `1030.5g, 406.5 g`D. `1145g, 234g`

Answer» Correct Answer - B
`Mn_(3)O_(4)` when heated with `AI` powder …………….
`3Mn_(3)O_(4)+8AI rarr 9 Mn+4AI_(2)O_(3)`
Minimum moles of `AI_(2)O_(3)` to be produced `= (612)/(102) = 6`
& minimum moles of `Mn` to be produced `= (825)/(55) = 15`
To produce 6 mole of `AI_(2)O_(3)`, mole required :
`n_(Mn_(3)O_(4) = (6xx3)/(4) = 4.5`
`n_(AI) = (8xx6)/(4) = 12`
To produce 15 moles of Mn, moles required :
`n_(Mn_(3)O_(4) = (15xx3)/(9) = 5`
`n_(AI) = (15xx8)/(9) = 13.33`
`:.` Minimum moles of reactant required :
`n_(Mn_(3)O_(4)) = 5& n_(AI) = 13.33`
`:. m_(Mn_(3)O_(4)) = 5xx229 = 1145g & m_(AI) = 13.33xx27 = 360 g`


Discussion

No Comment Found

Related InterviewSolutions