A student mixed 10.0 g of iron (III) chloride (FeCl3) with 8.0 g of sodium hydroxide (NaOH). If the theoretical yield of the reaction that occurs is 12.0 g of precipitate, what is the limiting reactant and what is the percent yield
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The student obtains 8.21 g of the product. The molar mass of Iron (III) Hydroxide is approximately 106.87 g/mol. Therefore, the theoretical yield of Iron (III) Hydroxide is 1 mole * 106.87 g/mol = 106.87 g. So, the theoretical yield of the product, Iron (III) Hydroxide, is 106.87 g. What is the percentage yield?
Jeremy H.
What is the theoretical yield (in g) of iron(III) hydroxide when 10.0 mL of 0.250 M NaOH and 15.0 mL of 0.120 M FeCl3 are mixed? Hint: This is a limiting reactant problem! 3 NaOH(aq) + FeCl3(aq) → Fe(OH)3(s) + 3 NaCl(aq)
David C.
The student obtains 8.21 g of the product. The molar mass of Iron (III) Hydroxide is approximately 106.87 g/mol. Therefore, the theoretical yield of Iron (III) Hydroxide is 1 mole * 106.87 g/mol = 106.87 g. So, the theoretical yield of the product, Iron (III) Hydroxide, is 106.87 g. . Using the actual yield of the product, calculate the moles of the product formed
Susan H.
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