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SOLUTION STOICHIOMETRY

SOLUTION STOICHIOMETRY. By, Sondra. What Is This?. Solution Stoichiometry is the method of predicting or analyzing the volume and concentration of solutions involved in a chemical reaction.

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SOLUTION STOICHIOMETRY

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  1. SOLUTION STOICHIOMETRY By, Sondra

  2. 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.

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

  4. 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

  5. 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/molNa+35.45g/molCl=58.44g/molNaCl N=450g/58.44g/mol N=7.7 moles

  6. 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.

  7. 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 C1V1=C2V2

  8. Missing Variable • If we put the values that we know into this equation we come up with: .026mol/L*300L=C2750L • We get 750L from adding 250L to the original 300L • Solve for C2 Sidenote: In this equation you may only have one unknown variable present.

  9. Solving • .026mol/L*300L=C2750L • Divide both sides by 750L • (.026mol/L*300L)/750L=C2 • Now calculate the brackets and divide by 750L • C2=.01mol/L

  10. Sample 2 • Water is added to 120L of 8.50mol/L NH3 till the concentration reached 2.80mol/L. How much water is needed to reach this concentration? • To complete this question we will use: C1V1=C2V2

  11. Missing Variable • For this we need to solve for V2. Input the known variables. • 8.50mol/L*120L=2.80mol/L*V2 • Divide both sides by 2.80mol/L and solve. • (8.50mol/L*120L)/2.80mol/L=V2 • V2=364L

  12. 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.

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