A50.0 ml sample of an aqueous h2so4 solution is titrated with a 0.375 m naoh solution. the equivalence point is reached with 62.5 ml of the base. the concentration of h2so4 is m.

0.188M

Explanation:

Given parameters:

Volume of H₂SO₄ = 50mL = 0.05L

Concentration of NaOH = 0.375M

Volume of base used = 62.5mL = 0.0625L

Unknown:

Concentration of acid used = ?

Solution:

In a neutralization reaction, an acid reacts with a base to produce salt and water.

For the reaction above, we write the complete balanced equation first:

       H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Now to find the unknown, we simply work from the parameters of the known species which the base.

We can find the number of mole of the reacting NaOH and proceed to solving the problem.

To find the number of moles, we use the expression below:

   Number of moles of NaOH = Molarity x volume of NaOH

                                                = 0.375 x 0.0625

                                               = 0.023mole

From the reaction equation:

  2 mole of NaoH neutralized 1 mole of the acid used

 0.023 mole of NaOH will neutralize = 0.012mole of acid

Using this value we can solve for the molarity of the acid:

     Molarity of acid =  

 Molarity of acid =   = 0.188M

8 carbon atoms forming single bonds with 18 hydrogen atoms; a triple bond exists between the third and fourth carbon atom from the right

explanation:

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.