Question

Consider the reaction 4NH3 + 5O2 4NO + 6H2O. In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. a) Which reactant is the limiting reagent? _________________ b) How many grams of NO are formed? _________________ c) How many grams of the excess reactant remains after the reaction?

Answer 1

In order to find the limiting reactant we will need to first set up the properly balance equation, which the question already provided us:

4 NH3 + 5 O2 -> 4 NO + 6 H2O

Now from the information in the question we have:

3.25 g of NH3

3.50 g of O2

We will need their molar mass in order to find how many moles of each compound we actually have:

O2 = 32 g/mol

NH3 = 17.03 g/mol

As we can see, in a regular reaction, we will have a 4:5 molar ratio between NH3 and O2, which means that for every 4 moles of NH3 we will need 5 moles of O2 in order to proceed in the reaction, not let's find the number of moles in NH3

17.03 g = 1 mol

3.25 g = x moles

x = 0.19 moles of NH3

Now we will use the molar ratio to find the number of moles of O2

4 NH3 = 5 O2

0.19 NH3 = x O2

x = 0.24 moles

Now we need to check if 0.24 moles of O2 is under the 3.50 grams of mass, if the mass is below 3.50, we have an excess of O2, therefore the limiting reactant will be NH3, if we have a mass above 3.50, this means that we actually have an excess of NH3, and the limiting reactant is O2

32 g = 1 mol

x grams = 0.24 moles

x = 7.68 grams, which is above 3.50, which means that the limiting reactant is O2 and we have an excess of NH3

A) Oxygen gas, O2

Now for letter B

We have to use the limiting reactant to find out how much is the mass of the product, so let's use O2, which has 3.50 g of mass and molar mass of 32 g/mol

32 g = 1 mol

3.50 g = x moles

x = 0.11 moles of O2

The molar ratio is 5:4, 5 moles of Oxygen gas for 4 moles of NO, this in a regular situation, but we have 0.11 moles of O2, for NO will be:

5 O2 = 4 NO

0.11 O2 = x NO

x = 0.09 moles of NO

Now we have to find the mass, we will use its molar mass, which is 30.01 g/mol

30.01 g = 1 mol

x grams = 0.09 moles

x = 2.7 grams of NO

b) 2.7 grams of NO

For letter C

We already know that the oxygen gas is the limiting reactant, and we have 0.11 moles of O2 in the reaction, let's find out how many grams are in excess in 3.25 g of NH3, again we are going to use molar ratio 4:5

4 NH3 = 5 O2

x NH3 = 0.11 O2

x = 0.09 moles of NH3

Using the molar mass of NH3 we will end up with

17.03 g = 1 mol

x grams = 0.09 moles

x = 1.53 grams of NH3

Now we only need to subtract this from 3.25 g

3.25 - 1.53

We have an excess of 1.72 grams of NH3

c) 1.72 grams of NH3