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Consider the following reaction: 2H2S(g)+3O2(g) yields 2SO2(g) +2H2O(g). If O2 was the...

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psty | eNoter

Posted February 4, 2011 at 12:57 AM via web

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Consider the following reaction: 2H2S(g)+3O2(g) yields 2SO2(g) +2H2O(g). If O2 was the excess reagent 8.3 mol of H2S were consumed and 137.1 g of water were collected after the reaction has gone to completion what is the percent yield of the reaction. Please show work

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ndnordic | High School Teacher | (Level 2) Associate Educator

Posted February 4, 2011 at 3:28 AM (Answer #1)

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To solve a problem like this, look at the coefficients of the balanced chemical equation.  In this case, it tells you that for every two moles of H2S you react, you will produce two moles of HOH.  That assumes a 100% yield.

In the given problem, since you started with 8.3 moles of H2S, you would expect 8.3 moles of HOH at 100% yield.

In your case, you produced 137.1 g of water.  Convert this to moles which means divide 137.1 by the formula mass of water (18.016 g/mole) .

137.1 g/ 18.016 g/mole = 7.61 moles of water.

To find % yield, divide the actual yield by the theoretical yield and multiply by 100.

% yield = 7.61/8.3 * 100 = 91.69%

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