How many milliliters of a 0.100 M NaOH solution are needed to neutralize 15.0 mL of 0.200 M H3PO4?

90 mL

Explanation:

Step 1:

Data obtained from the question. This include the following:

Volume of base, NaOH (Vb) =..?

Molarity of base, NaOH (Mb) = 0.1 M

Volume of acid, H3PO4 (Va) = 15 mL

Molarity of acid, H3PO4 (Ma) = 0.2 M

Step 2:

The balanced equation for the reaction. This is given below:

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

From the balanced equation above,

Mole ratio of the acid, H3PO4 (nA) = 1

Mole ratio of base, NaOH (nB) = 3

Step 3:

Determination of the volume of the base. The volume of the base needed for the reaction can be obtained as follow:

MaVa / MbVb = nA/nB

0.2 x 15 / 0.1 x Vb = 1/3

Cross multiply

0.1 x Vb = 0.2 x 15 x 3

Divide both side by 0.1

Vb = (0.2 x 15 x 3) / 0.1

Vb = 90 mL

Therefore, 90 mL of the base, NaOH is needed for the reaction.

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.

where is the following

explanation: