Question

A sample of 10.5 G of nitrogen reacts with 20.2 G of hydrogen to produce ammonia.3H2(g) + N2 —> 2NH3(g)a. What is the theoretical yield of ammonia?b. What is the limiting reactant?c. What is the excess reactant?d. How much excess reactant remains?e. If 4.1 G of ammonia is actually produced, what is the percent yield?

Answer 1

In order to answer this question, we need to set the properly balanced equation first:

3 H2 + N2 -> 2 NH3

This step will be always the first step in a question that involves stoichiometry, and from a properly balanced equation, we can take some informations, like the molar ratio between compounds.

3H2 = 1N2

3H2 = 2NH3

1N2 = 2NH3

Here we can see the molar ratio of the compounds, this means that for every 3 moles of H2 in the reaction, we will only need 1 mol of N2 in order to proceed the reaction. The same thing is true for H2 and NH3, but now the molar ratio will be 3:2, and N2:NH3 will be 1:2.

Now that we have the molar ratio for the reaction, we need to find the limiting and excess reactants.

We have:

10.5 grams of N2, molar mass is = 28g/mol

20.2 grams of H2, molar mass is = 2g/mol

B)

Let's see which one is in excess, starting with Nitrogen:

28g = 1 mol

10.5g = x moles

x = 0.375 moles of N2 in the reaction, and because of the molar ratio, we need 3 times more H2 than that, 0.375*3 = 1.125 moles of H2 if we have 0.375 moles of N2, let's check if we have this amount, or more than that of H2

C)

2g = 1 mol

20.2g = x moles

x = 10.1 moles of H2, this shows us that we have way more H2 than we actually need in order to react with N2, therefore the limiting reactant will be N2 and the excess reactant will be H2

D)

The excess will be the amount that we have of the reactant minus the amount we actually need:

20.2g - 2.25 g = 17.95 grams of excess

A)

Now we can find the theoretical yield of NH3, based on the amount of limiting reactant. The molar ratio of N2 and NH3 is 1:2, which means if we have 0.375 moles of N2, we will have 0.75 moles of NH3 as product, and by using its molar mass, 17g/mol, we can find the final mass:

17g = 1 mol

x grams = 0.75 moles of NH3

x = 12.75 grams of NH3 will be produced

E)

To find the percent yield, we will use the formula:

%yield = actual yield/percent yield

%yield = 4.1g/12.75g

%yield = 0.3216

The percent yield will be 32.16%

A. 12.75 grams

B. Nitrogen gas, N2

C. Hydrogen gas, H2

D. 17.95 grams

E. 32.16%