Stoichiometry in Solutions

Same method for solving problems but may need to use concentration or NIE (net ionic equation) to solve for the amount of product or reactant in question

Ex:

Oxalate ions (contained rhubarb leaves) are poisonous and can cause respiratory failure. To determine the concentration of oxalate ions in the leaves they were ground up and excess calcium chloride was added to precipitate calcium oxalate. A 238.6g leaf sample yielded 0.556g of calcium oxalate. What was the mass percent of oxalate ions in the leaves?

Givens: mass leaves = 238.6 g mCaC $_{2}$ O $_{4}$ = 0.556 g

Unknown:

Mass % C $_{2}$ O $_4$$^-$$^2$ = mass C $_{2}$ O $_4$$^-$$^2$ / mass leaves * 100%

Strategy: (not finished take notes from mr.hoffmans presentation)

nCaC $_{2}$ O $_{4}$ -> mCaC $_{2}$ O $_{4}$ / MCaC $_{2}$ O $_{4}$ = $\frac{0.556 g}{128.1 g / m o l}$
Rxn: C $_{2}$ O $_4$$^-$$^2$ (aq) + CaCl $_{2}$ -> CaC $_{2}$ O $_{4}$ (s) + 2 Cl $^{-}$ TIE: C $_{2}$ O $_4$$^-$$^2$ (aq) + Ca $^2$$^+$ (aq) + 2Cl $^{-}$ (aq) -> CaC $_{2}$ O $_{4}$ + 2 Cl $^{-}$

Example 2: Lead is a highly poisonous metal which must be removed from drinking water. If 30.0 mL of 0.075 M of sodium sulfate is added to 25.8 mL of 0.15 M lead (II) nitrate what is the mas of lead (II) sulfate that precipitates?

change volumes to liters because M = mol / Liters

Grade 11 Chemistry (Homework)

p. 410 # 1 - 10 p. 414 # 1 - 14 p. 417 # 11 - 20 p. 418 # 7 - 12 p. 420 # 21 - 30 p. 421 # 1 - 13

Ayush Garg

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