2. SOLUTION STOICHIOMETRY By, Sondra

3. What Is This? Solution Stoichiometry is the method of predicting or analyzing the volume and concentration of solutions involved in a chemical reaction. The major difference in solution stoichiometry from the general stoichiometric method is that the amount of volume and concentration are used as the conversion factors.

4. Calculating Stoichiometric Problems I. Write a balanced equation, along with all quantities and conversion factors. II. Convert the given measurements to its chemical amount using the appropriate conversion factors.(eg. Grams, moles…) III. Calculate the amount of the other substance using the mole ratio from your balanced equation IV. Convert the final amount to the quantity requested.

5. Sample 1 A student dissolves 450g of sodium chloride into 300L of water to make NaCl(H2O). What is the concentration of NaCl in the water? To solve this question we will be using 2 formulas N=M/m and C=N/V

6. Moles Since NaCl + H2O = NaCl(H2O) is already a balanced equation we can skip to moles. The number of moles of NaCl in a 450g sample is found as follows: 22.99g/mol Na +35.45g/mol Cl =58.44g/mol NaCl N=450g/58.44g/mol N=7.7 moles

7. Consentration Now that we know how many moles of NaCl we have we can continue to find concentration. This is done as follows: C=7.7mol/300L C=0.026mol/L So we now know that 450g of sodium chloride in 300L of water has a concentration of.026mol/L.

8. Volume To continue with our previous question what would the concentration of the solution be if we added 250L of water to the already 300L? To find this type of question we need the following formula C 1 V 1 =C 2 V 2

9. Missing Variable If we put the values that we know into this equation we come up with:.026mol/L*300L=C 2 750L We get 750L from adding 250L to the original 300L Solve for C 2 Sidenote: In this equation you may only have one unknown variable present.

10. Solving.026mol/L*300L=C 2 750L Divide both sides by 750L (.026mol/L*300L)/750L=C 2 Now calculate the brackets and divide by 750L C 2 =.01mol/L

11. Sample 2 Water is added to 120L of 8.50mol/L NH 3 till the concentration reached 2.80mol/L. How much water is needed to reach this concentration? To complete this question we will use: C 1 V 1 =C 2 V 2

12. Missing Variable For this we need to solve for V 2. Input the known variables. 8.50mol/L*120L=2.80mol/L*V 2 Divide both sides by 2.80mol/L and solve. (8.50mol/L*120L)/2.80mol/L=V 2 V 2 =364L

13. Conclusion In conclusion as long as you follow the steps and know those three formulas you will be able to figure out any solution question.