Assume 0.025 mol of N2 and 0.809 mol of H2 are present initially. 1. After compl

Question

Assume 0.025 mol of N2 and 0.809 mol of H2 are present initially. 1. After completing the reaction, how many miles of ammonia are produced? 2. How many miles of H2 are produced? 3. How many moles of N2 are produced? 4. What is the limiting reactant? (Nitrogen or hydrogen) itrogen and hydrogen combine at high temperature, in the presence N2(g)+3H2(g) 2NH,(g) Assume 0.250 mol of N2 and 0.809 mol of H2 are present initialy. Number After complete reaction, how many moles of ammonia are produced? mol Number How many moles of H2 remain? mol Number How many moles of N2 remain? mol What is the limiting reactant? nitrogen O hydrogen o hydrogen

Explanation / Answer

Solution :-

0.250 mol N2

0.809 mol H2

Lets calculate the moles of H2 needed to react with 0.250 mol N2

0.250 mol N2 * 3 mol H2 / 1 mol N2 = 0.750 mol H2

Moles of H2 needed are less than moles of H2 present

Therefore N2 is the limiting reactant

Now lets calculate the moles of NH3 produced

0.250 mol N2 * 2 mol NH3 / 1 mol N2 = 0.500 mol NH3 formed

Moles of H2 remain = 0.809 mol – 0.750 mol = 0.059 mol H2 remain

Moles of N2 remain = 0 mol because it is limiting reactant so used up completely .

Limiting reactant = N2 (nitrogen)

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