Question

hi someone please explain to me very detailed how to do this i kind of have an idea of how but im not rlly sure

Answer 1

Step 1 - Understanding what is excess and limiting reactants

Every reaction happens according to a fixed proportion. When one of the reactants has a mass that is bigger than needed (i.e., bigger than the required proportion), it will not react completely: a part of it will remain, which we thus call excess.

On the other hand, the other reactant will react according to the proportion. When it is over, the reaction stops, i.e., it limits the reaction, being thus called limiting.

Step 2 - Finding the limiting reactant for the given reaction

The given reaction is:

`N_2+3H_2\rightarrow2NH_3`

Let's read this reaction:

one mole of N2 reacts with three moles of H2

We can convert this to a proportion in grams by multiplying the number of moles of each reactant by its respective molar mass (28 g/mol for N2; 2 g/mol for H2):

`\begin{gathered} H_2\rightarrow3moles*2g/mol=6g \\ N_2\rightarrow1mole*28g/mol=28g \end{gathered}`

Therefore:

28g of N2 react with 6g of H2

We can use this proportion to determine the limiting reactant.

We have reacted 10g of N2 with 2.5g of H2. Let's see how much H2 is needed to completely react with 10g of N2:

`\begin{gathered} 28g\text{ of N2 react with -----6g of H2} \\ 10g\text{ of N2 would react with --- x} \\ \\ x=(60)/(28)=2.14g\text{ of H2} \end{gathered}`

We would need less H2 than what was actually reacted. Therefore, when 10g of N2 are completely consumed, only 2.14g will have reacted. H2 is thus in excess and N2 is the limiting reactant.

Answer: H2 is in excess and N2 is the limiting reactant