6. What volume of 6M Ba(OH)2 will be needed to neutralize 400.0 ml of a 2.0 M HCl solution?

133 mL

Explanation:

M= mol solute/ L solution

mol solute= M *L solution

mol solute = (2.0 M * 0.4 L)

mol solute= 0.8 mol HCl= mol H

We need the same amount of  OH moles to neutralize the solution

L solution = mol solute/M

L solution= 0.8 mol Ba(OH)2/ 6M

L solution=0.133 L=133 mL

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.

the correct option is c.

isotopes refers to different forms of the same element with the same atomic number but different mass number as a result of difference in their neutron numbers. isotopes have different relative atomic mass but share the same chemical properties.