Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the percent yield when 3.78 g of ammonium sulfate forms from the reaction of 2.51 g of sulfur trioxide, 1.55 g of ammonia, and 0.804 g of water. Can you please show step by step on how you got the answer thanks!
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First, the balanced equation:
1 SO3 + 2 NH3 + 1 H2O --> 1 (NH4)2SO4
Find the limiting reactant using molar masses and mole ratios
2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) = 0.03135 mol (NH4)2SO4
1.55 g NH3 (1 mol NH3 / 17.04 g NH3) (1 mol (NH4)2SO4 / 2 mol NH3) = 0.04548 mol (NH4)2SO4
0.804 g H2O (1 mol H2O / 18.02 g H2O) (1 mol (NH4)2SO4 / 1 mol H2O) = 0.04462 mol (NH4)2SO4
Therefore, the limiting reactant is sulfur trioxide.
Now find the theoretical yield of (NH4)2SO4.
2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) (132.16 g (NH4)2SO3 / 1 mol (NH4)2SO4) = 4.14 g (NH4)2SO4
Actual yield / Theoretical yield x 100% = Percent yield.
Therefore, percent yield = 91.3%
1 SO3 + 2 NH3 + 1 H2O --> 1 (NH4)2SO4
Find the limiting reactant using molar masses and mole ratios
2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) = 0.03135 mol (NH4)2SO4
1.55 g NH3 (1 mol NH3 / 17.04 g NH3) (1 mol (NH4)2SO4 / 2 mol NH3) = 0.04548 mol (NH4)2SO4
0.804 g H2O (1 mol H2O / 18.02 g H2O) (1 mol (NH4)2SO4 / 1 mol H2O) = 0.04462 mol (NH4)2SO4
Therefore, the limiting reactant is sulfur trioxide.
Now find the theoretical yield of (NH4)2SO4.
2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) (132.16 g (NH4)2SO3 / 1 mol (NH4)2SO4) = 4.14 g (NH4)2SO4
Actual yield / Theoretical yield x 100% = Percent yield.
Therefore, percent yield = 91.3%
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can you help me again with a similar problem. http://answers.yahoo.com/question/index;_ylt=AjoE9ta1SVUegy4iIjFvEUzsy6IX;_ylv=3?qid=20120618203357AAJpO9T for the mole ratio in that problem which one do i use?
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