oxygen gas for every molé of metiahe Iih the reacl6h IS 10 Fun to 8 When you com

Question

oxygen gas for every molé of metiahe Iih the reacl6h IS 10 Fun to 8 When you combust methane you need two moles of over. If you have five moles of methane and five moles of oxygen you can produce two moles of carbon dioxide and four is available, you do not have enough oxygen gas to produce more carbon dioxide. Thus you can conclude that oxygen is Part D Find the number of moles of water that can be formed if you have 190 mol of hydrogen gas and 90 mol of oxygen gas. Express your answer with the appropriate units. Hints ValueUnits Submit My Answers Give Up Ps

Explanation / Answer

the balanced reacrion will be

2h2(g) + o2(g) -----> 2h2o(l)

by seeing the stoichiometric coefficients of o2 and h2 ....we can say that

1 mol of o2 requires 2 moles of h2

so , 90 mol of o2 requires (90×2) moles of h2

so for the completion of reaaction we need 90 moles of o2 and 180 moles of h2....

hence , 10 moles of h2 will be left behind....so , o2 is the limiting reagent and h2 is the excess reagent....

Clearly , 1 mol of o2 produces 1 mol of water

So, 90 mol of o2 will produce 90 moles of water.....

So , moles of water produced is 90 moles....

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