Question

What's the best way to do stoichiometry

Answer 1

Solution:- The best way for doing stoichiometry is to write the balanced equation and then use dimensional analysis for solving the problem.

If the amount is given for only one of the reactant and says that other reactant is in excess then It's very easy to do the calculations.

Example:- 6.00 g of H₂ react with O₂ in excess to form H₂O. Calculate the amount of H₂O formed.

First write the balanced equation...

2H₂ + O₂ -----> 2H₂O

Start with the given one that is grams of hydrogen and convert them to moles. Then use the mol ratio and calculate the moles of water and finally convert these moles to grams by multiplying molar mass of water.

So, the steps would be like, grams of H₂ ---> moles of H₂ ---> moles of H₂O ----> grams of H₂O

6.00 g H₂ x (1 mol H₂/2.02 g H₂) x (2 mol H₂O/2 mol H₂) x (18.02 g H₂O/ 1mol H₂O)

Here we have used 2:2 mol as from balanced equation, 2 moles of H₂ form 2 moles of H₂O.

on doing calculations we get 53.5 g H₂O.

If moles of H₂ are given and we are asked to calculate the grams of H₂O then we would exclude the first ste which is grams of H₂ ----> moles of H₂.

If moles of H₂ are given and we are asked to calculate the moles of H₂O then we would exclude first step as well as as last step.

Now, if the grams are given for both the reactants then we need to find out the limiting reactant, It is the reactant that is use up totally and the formation of the product depends on this reactant.

The easy concept for this is, calculate the amount of the product starting with the amount of both the reactants and see which one of them gives us less amount of product. The limiting reactant is the one that gives less amount of product.

Example:- 5.00 g of H₂ react with 10.0 g of O₂ to form water. Calculate the amount of water that would form?

5.00 g H₂ x (1 mol H₂/2.02 g H₂) x (2 mol H₂O/2 mol H₂) x (18.02 g H₂O/ 1mol H₂O) = 44.6 g H₂O

Now let's do the calculations using the grams of O₂.

10.0 g O₂ x (1 mol O₂/32.0 g O₂) x ( 2 mol H₂O/ 1mol O₂) x (18.02 g H₂O/ 1 mol H₂O) = 11.3 g H₂O

Since O₂ gives less amount of water so it is the limiting reactant and we get 11.3 g of water.

If we are asked how many grams of H₂ are in excess then we could do this by calculating the grams of hyrogen used to react with given grams of oxygen. The calculations would be as...

10.0 g O₂ x (1 mol O₂/32.0 g O₂) x (2 mol H₂/1mol O₂) x (2.02 g H₂/1 mol H₂) = 1.26 g H₂

excess amount of H₂ = 5.00 g - 1.26 g = 3.74 g

If asked to calculate the percent yield then actual yield must be given and the amount of prouct that we calculate doing the calculations in above examples is the theoretical yield.

The formula is, percent yield = (actual/theoretical)*100