Question

In the air nitrogen monoxide can react with oxygen gas to form nitrogen dioxide gas, a contributor to photochemical smog. What is the theoretical yield of nitrogen dioxide gas that can be generated from the reaction of 22.0 g of nitrogen monoxide gas with 22.0 g of oxygen gas?Equation: ?Limiting Reactant: ?Excess: ?Grams of target product produced: ?

Answer 1

In order to solve this question we need to first:

Set up the properly balanced equation, which is:

2 NO + O2 -> 2 NO2

One important information is that the molar ratio between NO and O2 is 2:1, which means that for every 2 moles of NO, we will need 1 mol of O2 in order to proceed with the reaction

Now we are given:

22.0 grams of NO

22.0 grams of O2

So now we need to find which compound is in excess and which one is the limiting reactant, let's start with NO, which has a molar mass of 30.01g/mol

30.01 g = 1 mol of NO

22.0 g = x moles of NO

x = 0.73 of moles of NO in 22 grams

According to the molar ratio, if we have 0.73 moles of NO, we will need half of it of O2, which will be 0.36 moles of O2, but we don't know if we have this number of moles or more, let's see:

molar mass for O2 is 32g/mol

32g = 1 mol

x grams = 0.36 moles

x = 11.5 grams of O2 is needed, but we have 22.0 grams available, therefore we have O2 in excess and the limiting reactant will be NO

Now that we have the limiting reactant, we can use it to find out how much will be produced

The molar ratio between NO and NO2 will be 2:2, same number of moles for both compounds, therefore if we have 0.73 moles of NO, we will also have 0.73 moles of NO2 being produced

Using the molar mass of NO2, which is 46g/mol, we can find the final mass

46g = 1 mol

x grams = 0.73 moles of NO2

x = 33.6 grams of NO2 will be produced and it will be the theoretical yield