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CHEMISTRY 101 Lecture Note, Lecture notes of Chemistry

Dimensional Analysis Stoichiometry with solutions

Typology: Lecture notes

2021/2022

Uploaded on 09/07/2023

MadeleineJames
MadeleineJames 🇬🇧

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In this example, we use dimensional analysis to solve a problem involving stoichiometry with
solutions. Our problem reads, "What volume of 0.15 molar sodium phosphate is needed to
completely react with 22.5 milliliters of 0.24 molar nickel two chloride, according to the reaction
below?"
We'll start our dimensional analysis with 22.5 milliliters of nickel two chloride solution. We can
convert from milliliters to liters of solution. Then we can use the molarity of nickel two chloride
to convert from liters of solution to moles of nickel two chloride. We have 0.24 moles of nickel
two chloride in each liter of solution. Then we can use the coefficients from the balanced
chemical equation to convert from moles of nickel two chloride to moles of sodium phosphate.
We'll use the concentration of sodium phosphate to convert from moles of solute to liters of
sodium phosphate solution. And then we can convert back to milliliters of solution. Completing
this calculation, we should get 24 milliliters of sodium phosphate solution.

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In this example, we use dimensional analysis to solve a problem involving stoichiometry with solutions. Our problem reads, "What volume of 0.15 molar sodium phosphate is needed to completely react with 22.5 milliliters of 0.24 molar nickel two chloride, according to the reaction below?" We'll start our dimensional analysis with 22.5 milliliters of nickel two chloride solution. We can convert from milliliters to liters of solution. Then we can use the molarity of nickel two chloride to convert from liters of solution to moles of nickel two chloride. We have 0.24 moles of nickel two chloride in each liter of solution. Then we can use the coefficients from the balanced chemical equation to convert from moles of nickel two chloride to moles of sodium phosphate. We'll use the concentration of sodium phosphate to convert from moles of solute to liters of sodium phosphate solution. And then we can convert back to milliliters of solution. Completing this calculation, we should get 24 milliliters of sodium phosphate solution.