For a reaction, `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`, identify dihydrogen `(H_(2))` as a limiting reagent in the following reaction mixtures.

Options

A. 28 g of `N_(2) + 6g` of `H_(2)`
B. 56 g of `N_(2) + 10 g` of `H_(2)`
C. 14 g of `N_(2) + 4g H_(2)`
D. 35g of `N_(2) + 8g` of `H_(2)`

Answer

B

Explanation

from given eq :
`{:(N_(2)(g),+,3H_(2)(g),rarr,2NH_(3)(g)),("1 mole",,"3 mole",,),(1 xx 28 gm,,3 xx 2gm,,),("= 28 gm",,"= 6 gm",,):}`
For option -1 `because 6 gm H_(2)` reacts `bar(c )` 28 gm `N_(2)` i.e. both reactants utilized completely.
for option `because gm H_(2)` reacts `bar(c )` 28 gm `N_(2) :. 10 gm H_(2)` reacts `(28)/(6) xx 10 gm N_(2)`
`= 46.67 gm N_(2)` required, however amount. `N_(2)` of given = 56 gm
i.e. `N_(2)` Dresent in excess.
So, `H_(2)` utilizes completely & hence limiting reagent.
for option - 3 `because 6 gm H_(2)` reacts `bar(c )` 28 gm `N_(2) :. 4 gm H_(2)` reacts `bar(c ) (28 xx 4)/(6) gm N_(2)`
= 18.67 gm `N_(2)` required. But `N_(2)` given = 14 gm. `:. N_(2)` i L.R.
for option - 4 `because 6gm H_(2)` reacts `bar(c )` 28 gm `N_(2)`
`because 6 gm H_(2)` reacts `bar(c ) (28 xx 8)/(6) gm N_(2) = 37.33 gm N_(2)` required by `N_(2)` given = 35 gm.