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### Topic: stoichiometry: percentage yield  (Read 824 times)

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#### 073735557

• Very New Member
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• Mole Snacks: +0/-0 ##### stoichiometry: percentage yield
« on: November 17, 2019, 10:59:06 AM »
The percentage yield of the following reaction is consistently 75.2%.
CH4 + 4S -----> CS2 + 2 (H2S)
What mass of sulphur is required to produce 50.0 g of hydrogen sulphide (H2S) if methane (CH4) is in excess?

i know the answer is 125, but i dont understand how to get there. i mainly understand stoichiometry through steps, so pls help me my stoichio test is tomorrow

and how would i usually approach any question involving percentage yield? what should i look for that will change my equations?

#### chenbeier

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• Gender:  ##### Re: stoichiometry: percentage yield
« Reply #1 on: November 17, 2019, 11:12:54 AM »
You calculate the moles of Hydrogensulfide. 50 g/34 g/mol = 1,47 mol
In the chemical equation you see sulphur to hydrogensulfide 4:2 = 2:1
The moles multiplied by 2 and the molar mass of sulfur gives 94,1 g
The yield is 75,2%. Means you need excess sulphur 94,1g/0.752 = 125,2 g