N2(g)+3H2(g)⟹2NH3(g)
moles of Nitrogen here =287.19×109=2.56×108
moles of Hydrogen here =21.19×109=5.95×108
here for every mole of nitrogen 3 moles of hydrogen is required and 2 moles of ammonia is formed
clearly from the above mole quantities we can see that hydrogen is the limiting reagent here
x=35.95×108=1.98×108
moles of ammonia formed =2x=2×1.98×108=3.96×108
mass of ammonia formed =17×3.96×108=6.73×109grams
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