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ructure.com/courses/2213886/quizzes/4307140/take for Teache D Question 6 12.5 pts N2 (g)+3H2 (8) 2 NH3 (8) The equation above is the equation for the Haber process. In a certain reaction, you start with 3.0 moles of nitrogen and 5.0 moles of hydrogen, which is the excess reactant? O neither O hydrogen O nitrogen 12.5 pts DQuestion 7 N2 (8)+3H2 (g) >2 NH3 (g) The equation above is the equation for the Haber process In a certain reaction, you start with 3.0 moles of nitrogen and 5.0 moles of hydrogen, which molecule is the limiting reagent?

Explanation / Answer

In the reaction N2 (g) + 3 H2 --> 2 NH3 (g)

From the equation one can find that 1 mole of N2 reacts with 3 moles of H2 and form 2 moles NH3, so for every 1 mole of N2, it requires 3 moles of H2 for complete reaction.

Q6. If you start with 3.0 moles of N2 and 5.0 moles of H2, N2 will be in excess, because 3.0 moles of N2 required 3.0 x3 = 9.0 moles of H2 but only 5.0 moles of H2 are available. So it can use a maximum of 5/3 = 1.6667 moles of N2, means N2 is excess.

Q7. The limiting reagent is the reagent for which the reactants available is lease amount. For 3.0 moles of N2 it required 9.0 moles of H2 for 100 % conversion, but only 5.0 moles of H2 are available. Out of 3 moles of N2 only 1.6667 moles of N2 are utilized, as during this much time all H2 get exhausted. Means H2 is the rate-limiting reagent.

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