If a mixture of 39.4 g CH4 and 72.1 g S reacts, how many grams of H2S are produced? CH4(g) + 4 S(g) → CS2(g) + 2 H2S(g)?
I keep trying this problem but it says it's wrong, I got 78.8g and 36.05g and you're supposed to take the lesser value and use the correct number of sig figs. Help!
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you have to find the limiting reagent first ....
no.of moles of CH4 = 39.4/molar mass = 39.4/16 = 2.463
no.of moles of S = 72.1 /32 = 2.253
according to reaction 1 mole of CH4 reacts with 4 moles of S
so 2.463 moles of CH4 will react with 2.463 X 4 = 9.852 moles of S but no.of moles of S given to you in the question is 2.253 which is less than what is needed so S will be the limitng reagent and amount of products formed will depend on amount of S only
according to reaction 4 moles of S gives 2 moles of H2S
so 2.253 moles of S will give 2/4 X 2.253 = 1.127 moles of H2S
molar mass of H2S = 34 g/mole
it means that 1 mole of H2S = 34 g
so 1.127 mole of H2S = 1.127 X 34 = 38.318 g of H2S