A sample of vinegar containing 0.100 mol of ethanoic acid has a concentration of 0.125M of ethanoic acid. What is the volume of the solution.

0.8 L

We are given:

Molarity of Ethanoic Acid in vinegar = 0.125 M

Number of moles of Ethanoic Acid = 0.1 mole

Finding the volume of the solution:

We know that the molarity of a solution is defined as:

Molarity = n / V   (where n is the number of moles and V is the volume in L)

replacing the variables with the given values:

0.125 = 0.1 / V

Multiplying by V on both the sides

0.125V = 0.1

Dividing both sides by 0.125

V = 0.1 / 0.125

V = 0.8 L

Therefore, the volume of the solution is 0.8 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.

i think it's c. i'm not sure but i think so.