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Title: Write a balanced equation for the reaction of nitrogen gas and hydrogen gas to produce Ammonia gas (NH3)? Post by: bugaian on Apr 8, 2012 B) If 212.5g of ammonia gas is produced, how many molecules of hydrogen gas must have reacted with nitrogen?
I need full steps...I don't get it... Title: Write a balanced equation for the reaction of nitrogen gas and hydrogen gas to produce Ammonia gas (NH3)? Post by: FISH0818 on Apr 8, 2012 N2 + 3H2 --> 2NH3
number of moles of NH3 = 212.5 / (14.0+3*1.00) = 12.5 from balanced equation, 1 mole of N2 and 3 moles of H2 will produce 2 moles of NH3. so now the 12.5 moles NH3 is actually formed by: 12.5/2 = 6.25 moles of N2 12.5/2*3 = 18.75 moles of H2 so number of molecules of H2 = 6.25 * Avogadro's number = 3.76*10^24 number of molecules of N2 = 18.75* Avogadro's number 1.13*10^25 |