When 24.68 grams of lithium nitrate (LiNO3) are produced, how many moles of lithium chloride (LiCI) reacted ?

0.3 moles of lithium chloride (LiCI) reacted to produce 24.68 grams of lithium nitrate (LiNO3).

Explanation:

Balanced equation for the reaction is:

2LiCl  + Pb(NO3)2 ⇒2 LiNO3 + PbCl2

Data given:

mass of LiNO3 produced = 24.68 grams

moles of LiCl reacted= ?

atomic mass of 1 mole of LiNO3 = 68.946 Grams

Number of moles =

putting the values in the above equation:

= 0.3 moles

from the reaction,

2 moles of LiCl reacted to form 2 moles of LiNO3

so, x moles of LiCl reacted to form 0.3 moles of LiNO3

=

x = 0.3 moles of LiCl will react to form 0.3 moles of LiNO3.

ur answer would be b hope i let me know if it is right

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.