Answer:
Percent Yield (Na₂CO₃) = 96%
Percent Yield (H₂CO₃) = 165%
This product is likely Na₂CO₃ because the mass is the closest to the given value. Also, its percent yield is the closest to 100%.
Explanation:
To find the answers, you need to (1) calculate the molar masses of each reactant and product, then (2) convert grams NaHCO₃ to grams products (using molar masses and mole-to-mole ratio from balanced equation), and then (3) calculate the percent yield.
(Step 1)
Molar Mass (NaHCO₃):
22.990 g/mol + 1.008 g/mol + 12.011 g/mol + 3(15.998 g/mol)
Molar Mass (NaHCO₃): 84.003 g/mol
Molar Mass (Na₂CO₃): 2(22.990 g/mol) + 12.011 g/mol + 3(15.998 g/mol)
Molar Mass (Na₂CO₃): 105.985 g/mol
Molar Mass (H₂CO₃): 2(1.008 g/mol) + 12.011 g/mol + 3(15.998 g/mol)
Molar Mass (H₂CO₃): 62.021 g/mol
(Step 2)
2 NaHCO₃ -----> 1 Na₂CO₃ + 1 H₂CO₃
3.55 g NaHCO₃ 1 mole 1 mole Na₂CO₃ 105.985 g
------------------------- x ------------------ x --------------------------- x ------------------ =
84.003 g 2 moles NaHCO₃ 1 mole
= 2.24 g Na₂CO₃
3.55 g NaHCO₃ 1 mole 1 mole H₂CO₃ 62.021 g
------------------------ x ----------------- x --------------------------- x ----------------- =
84.003 g 2 moles NaHCO₃ 1 mole
= 1.31 g H₂CO₃
<u>This product is likely Na₂CO₃ because the mass is the closest to the given value.</u>
(Step 3)
Actual Yield
Percent Yield = ------------------------------- x 100%
Theoretical Yield
2.16 g Na₂CO₃
Percent Yield = ------------------------------- x 100%
2.24 g Na₂CO₃
Percent Yield (Na₂CO₃) = 96%
2.16 g H₂CO₃
Percent Yield = ------------------------------ x 100%
1.31 g H₂CO₃
Percent Yield (H₂CO₃) = 165%
<u>The percent yield of Na₂CO₃ is closer to 100% than H₂CO₃. This also confirms that the product weighing 2.16 grams is likely Na₂CO₃.</u>
