Stoichiometry+Calculations+Involving+Molar+Concentration

= Stoichiometry Calculations Involving Molar Concentration =

(Page129-131 of the textbook)

===**Stoichiometry calculations are based on the relationship between the moles of one chemical and the moles of another chemical that in the same reaction.**=== **concentration ** (of a substance in solution )--- the amount of the substance which exists in a given volume of the solution

c=n/V
=n=c ∙V = =M(molarity)=mol/L =

also, we may use these:
= 22.4 L/ 1 mole = = 1 mole/ 22.4L =

**Improtant**: "22.4L" is used only if the substance is a **GAS** and the key phrase "**at STP**" is mentioned along with the volume.
**Rememberalways use known chemical to solve the unknown chemical**. It means : Known**→moles of know** n**→moles of known:moles of unknown→moles ** of unkown**→unknown**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 140%;">E.x. A student wants to put 50.0L of hydrogen gas at STP into a plastic bag by reacting excess aluminum metal (Al) with 3.00M sodium hydroxide solution (NaOH) according to the reaction: <span style="font-family: 'Times New Roman',Times,serif; font-size: 110%;">**2Al(s) + 2NaOH(aq) + 2H2O(l)** → **<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">2NaAlO 2 <span style="font-family: 'Times New Roman',Times,serif;"> (aq) + 3H 2 (g) ** <span style="font-family: 'Times New Roman',Times,serif; font-size: 140%;"> What volume of NaOH solution is required?

TO solve this problem, we can make a plan to think about the steps of how to solve it, like this:
 * 1) find out the moles of H 2 to find out the moles of NaOH
 * 2) use moles of NaOH to find out the volume of NaOH solution

<span style="color: #ff00ff; font-family: 'Comic Sans MS',cursive; font-size: 120%;">This is just my way to solve it. It follows the rules that I'd already told <span style="color: #ff0000; font-family: 'Comic Sans MS',cursive; font-size: 120%;">(known--moles of known--moles of known/moles of unknown--moles of unknown--unknown ).

<span style="color: #6000ff; font-family: 'Comic Sans MS',cursive; font-size: 130%;">Next, do you remember "Titration" and "Equivalence point" ?

"A titration is used to determine the unknown concentration of a solution." [Reference of the picture:
 * <span style="font-family: 'Berlin Sans FB Demi','sans-serif'; font-size: 12pt;">[]] **

"<span style="color: #934be7; font-family: 'Comic Sans MS',cursive; font-size: 140%;">Titration is a process in which a measured amount of a solution with a known concentration is reacted with a known volume of another solution unknown concentration, until a desired <span style="color: #934be7; font-family: 'Comic Sans MS',cursive; font-size: 140%;">equivalenve point is reached (neutralization)." [Reference: from the notes of Unit 7 Lesson 3]

It means that when we have a solution with known volumn and known concentration, we can measure the unknown concentration of another solution that with a known volume. Two kinds of solution are mixed and they react with each other, when the PH around 7, the equivalence point is reached, and the process of this is called <span style="color: #934be7; font-family: 'Comic Sans MS',cursive; font-size: 140%;">Titration.

The <span style="color: #934be7; font-family: 'Comic Sans MS',cursive; font-size: 140%;">Equivalence Point (or Stoichiometric Point) is just the point that "in a titration where the ratio of the moles of each substance exactly equals the ratio of the coefficients of the substance in the balanced reaction equation". [from the textbook page 130]

Let's see the examples that in the textbook page 130

<span style="font-family: 'Arial','sans-serif'; font-size: 12pt;">When H2SO4 reacts completely with NaOH, H2SO4 + 2NaOH **<span style="font-family: 宋体; font-size: 14pt;">→ **<span style="font-family: 'Arial','sans-serif'; font-size: 12pt;">Na2SO4 + 2H2O Assume there is 0.0250mol of H2SO4 in a beaker. By using a burette to add NaOH solution slowly, when 0.0500mol of NaOH has been added, the equivalence point has been reached, the titration process is stopped.

<span style="color: #ce2ee0; font-family: 'Comic Sans MS',cursive; font-size: 120%;">One More Example A 10.00 mL sample of phosphoric acid solution is titrated with a 0.200M NaOH solution. The endpoint is reached when 17.03mL of NaOH is added. Calculate the concentration of the H3PO4 solution. H3PO4 + 2NaOH -- > Na2HPO4 + 2H2O

<span style="color: #0070c0; font-family: 'Cambria','serif'; font-size: 16pt;">Solution: <span style="color: #0070c0; font-family: 'Cambria','serif'; font-size: 16pt;">moles of NaOH= n =c **<span style="font-family: 'Cambria','serif'; font-size: 14pt;"> ∙ **<span style="color: #0070c0; font-family: 'Cambria','serif'; font-size: 14pt;">V= 0.200mol/mL **<span style="font-family: 'Cambria','serif'; font-size: 14pt;">∙ **<span style="color: #0070c0; font-family: 'Cambria','serif'; font-size: 14pt;">17.03mL= 3.406mol <span style="color: #0070c0; font-family: 'Cambria','serif'; font-size: 14pt;">moles of H3PO4= 3.406mol NaOH  **<span style="font-family: 'Cambria','serif'; font-size: 14pt;">∙ **<span style="color: #0070c0; font-family: 'Cambria','serif'; font-size: 14pt;">1 mol  H3PO4/2 mol NaOH = 1.703mol [H3PO4]=1.703mol / 10.00mL=0.1703M