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If you started with 0.183 mol of N2 and 0.549 mol of H2, and they completely reacted in the reaction vessel, determine the total moles of gas particles (n) there are during the initial and final conditions. Additionally, determine the ratio of the number of gas particles in the products to that of the reactants, then complete the statements below.

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Answer:

Initial we have 0.183 + 3*0.183 = 0.732 moles of gas (those will all be consumed).

In the products we have 0.366 moles of gas

Explanation:

Step 1: Data given

Number of moles of N2 = 0.183 mol

Number of moles of H2 = 0.549 mol

Step 2: The balanced equation:

N2 + 3H2 → 2NH3

Step 3: Calculate the limiting reactant

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

There is no limiting reactant. But will be completely consumed

Step 4: Calculate moles of NH3

For 1 mol of N2 we have 2 moles of NH3 produced

For 0.183 moles of N2 we have 2*0.183 = 0.366 moles of NH3 produced

Initial we have 0.183 + 3*0.183 = 0.732 moles of gas (those will all be consumed).

In the products we have 0.366 moles of gas

nangelina6766

Answer:

Initially, there are 0.732 mol of gas particles. After the reaction is complete, there are 0.366 mol of gas particles. Therefore, the ratio of product to reactant prticles is 5.0

Explanation:

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