Which of the following statements is/are correct

Options

A. The reaction.
`Bi + 4HNO_(3) + 3H_(2) O rarr Bi (NO_(3))_(3). 5H_(2) O + NO`
`2.09 g` of `Bi` in `HNO_(3)` produces `48.5g` of bismuth nitrate.
(Atomic weight `Bi = 209 g, Mw "of" Bi (NO_(3))_(3).5H_(2)O = 485 g)`
B. `4.0 g` of 63% `HNO_(3)` by mass is required to react with `2.09 g` of `Bi`
C. The volume of `NO` gas produced at `STP` (1 bar `273 K`) is `0.227 L`
D. The volume of `NO` gas prodcued at `SATP` (1 bar, `298K`) is `0.247 L`.

Answer

B, C, D

Explanation

a. Mole of `Bi = (2.09)/(209) = 0.01 "mol"`
Weight of bismuth nitrate `= (0.01 "mol" Bi)`
`((1 "mol" Bi (NO_(3))_(3) .5H_(2) O)/("mol" Bi))`
`((485 g Bi (NO_(3))_(3) 5H_(2)O)/("mol" Bi(NO_(3))_(3) 5H_(2)O))`
`= 0.01 xx 484 = 4.85 g`
Hence, (a) is wrong.
b. Weight of `HNO_(3) = (0.01 "mol" Bi)`
`((4 "mol" HNO_(3))/("mol" Bi))((63 g HNO_(3))/("mol" HNO_(3)))`
`((100 g "solution")/(63 g HNO_(3)))`
`= (0.01 xx 4 xx 63 xx 100)/(63) = 4 g HNO_(3)`
c. Molar volume of gas at `STP` (1 bar, `273 K`) = 22.7 L`
`:. Volume of `NO` gas `= 0.01 xx 22.7 = 0.227 L`
d. Molar volume of gas at `SATP` (1 bar, `298 K ) = 24.7 L`
`:.` Volume of `NO` gas `= 0.01 xx 24.7 xx 24.7 = 0.24 L`