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SOLUTION STOICHIOMETRY

Solution stoichiometry is a method used to predict and analyze the volume and concentration of solutions in chemical reactions. Unlike general stoichiometry, solution stoichiometry focuses on volume and concentration as conversion factors. This guide outlines the steps for calculating stoichiometric problems, including writing balanced equations, converting measurements to chemical amounts, and using mole ratios. Detailed examples illustrate calculating concentrations and determining volumes after adding substances. Mastering the three key formulas will enable you to solve any solution-related question with ease.

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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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