What is the molarity of 2.34 mol of calcium carbonate in 1860 ml of solution?
a. 0.79 mol/l
b. 1.26 mol/l
c. 2.34 mol/l
d. 4.35 mol/l

option b= 1.26  mol/L

Explanation:

Given data:

moles = 2.34 mol

volume = 1860 mL

Solution:

molarity = moles of solute/ volume in L

First of all we will convert the mL in L,

1860 mL / 1000 = 1.86 L

now we will put the values in formula,

molarity = 2.34 mol / 1.86 L

molarity = 1.26  mol/L

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.