Consider the reaction between hydrogen gas and nitrogen gas to form ammonia: 3 H

Question

Consider the reaction between hydrogen gas and nitrogen gas to form ammonia:
3 H2(g) + N2(g) => 2 NH3(g)
How many grams of ammonia could be produced by the reaction of 6.00 liters of hydrogen with 4.00 liters of nitrogen at STP?
The answer is 3.04 but can anyone explain to me how to get this answer? Thanks!

Explanation / Answer

3 H2(g) + N2(g) => 2 NH3(g) 6-3X-----;4-X-------;2X 3X = 6 => X=2 therefore, H2 is the limiting reagent NH3 formed = 2X = 4 liters(vol. is directly proportional to moles) PV = nRT => 101315 * 0.004 = n* 8.314 * 273 => n = 0.1786 moles m = 0.1786 * 17 = 3.04 g

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