Nitric acid is produced from `NH_(3)` in the following three steps,
(I) `4NH_(3) (g) + 5O_(2) (g) to 4NO + 6H_(2)O (I)`
`(II) 2NO(g) + O_(2)(g) to 2NO_(2)(g)`
(III) `3NO_(2)(g) + H_(2)O(I) to 2HNO_(3) (aq) + NO(g)`
% yield of (I), (II) and (III) are 40%, 60% and 70% respectively, then what volume of `NH_3` (g) at 1 atm and 0°C is required to produce 1075 g `HNO_3` ?

Options

A. 3413 L
B. 3500 L
C. 6826 L
D. 1750 L

Answer

A

Explanation

Number of moles of `NO_2` required = `1075/63 xx 3/2 xx 1/0.7 = 35.56` mol
Number of moles of NO required = `36.56/0.60`
Number of moles of `NH_3` required
`=36.56/0.6 xx 1/0.40 = 152.325`
Volume of `NH_(3)` at STP required `=152.325 xx 22.4 =~ 3413` L