A fertilizer manufacturer uses the reaction between hydrogen and nitrogen to cre

Question

A fertilizer manufacturer uses the reaction between hydrogen and nitrogen to create ammonia (assume no other reactants or products). If 9.5x10^3 moles of hydrogen and 1.7 x 10^3 moles of nitrogen react at 277k and 1.25 atm, what volume of ammonia is produced?
A fertilizer manufacturer uses the reaction between hydrogen and nitrogen to create ammonia (assume no other reactants or products). If 9.5x10^3 moles of hydrogen and 1.7 x 10^3 moles of nitrogen react at 277k and 1.25 atm, what volume of ammonia is produced?

Explanation / Answer

The chemical reaction here is,

N2 + 3H2 ---> 2NH3

So we have 1 mole of N2 reacting with 3 moles of H2 to form 2 moles of NH3

When NH3 is formed starting with 1.7 x 10^3 moles of N2 = 3.4 x 10^3 moles x 17 g/mol = 5.78 x 10^4 g NH3

when starting with H2 we would get = 9.5 x 10^3 x 2/3 = 6.33 x 10^3 mols x 17 g/mol = 1.08 x 10^5 g of NH3

So, here N2 is the limiting reagent

moles of NH3 formed would be = 3.4 x 10^3 mols

volume of NH3 formed would be = nRT/P

= 3.4 x 10^3 x 0.08206 x 277/1.25

= 6.18 x 10^4 L

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