Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g ) ⟶ 2 NH 3 ( g ) Assume 0.140 mol N2 and 0.434 mol H 2 are present initially. After complete reaction, how many moles of ammonia are produced

After complete reaction, 0.280 moles of ammonia are produced

Explanation:

Step 1: Data given

Number of moles N2 = 0.140 moles

Number of moles H2 = 0.434 moles

Step 2: The balanced equation

N2(g) + 3H2 (g) ⟶ 2NH3 (g)

Step 3: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.140 moles).

H2 is in excess. There will react 3*0.140 = 0.420 moles

There will remain 0.434 - 0.420 = 0.014 moles

Step 4: Calculate moles NH3

For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3

After complete reaction, 0.280 moles of ammonia are produced

a:

by the combination of an unlimited number of elements

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