underset20 gN 2( g )+ underset5 g3 H 2( g ) leftharpoons 2 NH 3( g ) Consider the above reaction, the limiting reagent of the reaction and number of moles of NH 3 formed respectively are:

Question

underset20 gN 2( g )+ underset5 g3 H 2( g ) leftharpoons 2 NH 3( g ) Consider the above reaction, the limiting reagent of the reaction and number of moles of NH 3 formed respectively are: (A) H 2, 1.42 moles (B) H 2, 0.71 moles (C) N 2, 1.42 moles (D) N 2, 0.71 moles

Tardigrade Admin · Accepted Answer

undersetW 2=20 gN 2( g )+ underset5 g3 H 2( g ) leftharpoons 2 NH 3( g ) n =(20/28) (5/2) Stoichiometric Amount: N 2 arrow (20 / 28/1)=(20/28) H 2 arrow (5 / 2/3)=(5/6) ∴ N 2 is the Limiting Reagent. ∴ n ( NH 3) =2 × n ( N 2)=2 × (20/28) =1.42

Q. $20 g N_{2 (g)} + 5 g 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}$ Consider the above reaction, the limiting reagent of the reaction and number of moles of $N H_{3}$ formed respectively are:

$H_{2}, 1.42$ moles

$H_{2}, 0.71$ moles

$N_{2}, 1.42$ moles

$N_{2}, 0.71$ moles

Q. 20gN2(g)​​+5g3H2(g)​​⇌2NH3(g)​ Consider the above reaction, the limiting reagent of the reaction and number of moles of NH3​ formed respectively are:

H2​,1.42 moles

H2​,0.71 moles

N2​,1.42 moles

N2​,0.71 moles

W2​=20gN2​(g)​+5g3H2​(g)​⇌2NH3​(g) n=2820​25​ Stoichiometric Amount: N2​→120/28​=2820​H2​→35/2​=65​ ∴N2​ is the Limiting Reagent. ∴n(NH3​)=2×n(N2​)=2×2820​ =1.42

Q. $20 g N_{2 (g)} + 5 g 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}$ Consider the above reaction, the limiting reagent of the reaction and number of moles of $N H_{3}$ formed respectively are:

$H_{2}, 1.42$ moles

$H_{2}, 0.71$ moles

$N_{2}, 1.42$ moles

$N_{2}, 0.71$ moles