Chemistry, 17.06.2021 06:30 help1219

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.

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Answer from: mjlc2099

Explanation:

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!

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%

hoppe it helps

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As cooking food that has to go in the oven

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the percent...

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