Question

You react 10.0 g of hydrogen gas with 60.0 g of oxygen gas to form water. Determine the amount of water formed and the amount of reactant in excess (both in grams) after the reaction is complete.

Answer 1

Geaction between rydrogen and oxygen to form water is as follows 2+299 + O2(g) ->2H20601 Given mass of h₂ = 10.09 man 42 = 60.08 We know that, molar mam 2 2 9 mal molar mass of o = 32g jmal moles of 2 = mass of 12 - 10.0g = [5.oomall Melon mass of My ?9 mel 9) Mel 60.09 12:males 0,- Masso malat mass of o, 30g 1.Ss mol we will find dinining acacrant in the section. wt know that, limiting racoar is the reachane which has least walues moles divided by stochrome- the coefficient males for nzi Stoc'omeni 2 = 5:00 = 2.50 for Oz moles 1. _ Stochiometry ne T = 288 = 1.ssmal O₂ hat least value of moles divided by stochiometry Of is the limiting reason D. Scanned with CamScanner

Also, by sto chiometry of reaction I more o reacts with a mole of give a mole of H₂o. the to . 1.88 mole olo, reacts with 2x1.88 mole of 4₂ so give 2x1.88 mole of n₂o. i mole of I reached = 3.76 mol mole of deft= 5.00-3.76 = 11.24 mol ) moles of 120 formed = 2x1.883.06 mol we know that molar mass of ₂0= 1891mol i. mass of 120= moles of h2o x molar mass of h2o mass of M20- 3.76 med X: 189) mal =167.79 mass of ₂ = moles of heeft x molar mass of 2 mass of 12 = 1.24 malx agnol mass 04 * = 12:48 9 ] cs Scanned with CamScanner
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