Question

Given the reaction $2{C_3}{H_7}OH + 9{O_2} \to C{O_2} + 8{H_2}O$ , how many moles of oxygen are needed to react with $3.40mol$ ${C_3}{H_7}OH$ ?

Answer 1

First of all we have to check that whether the equation is balanced, if not we have to balance it first to proceed to the calculation. Because from the correct balanced equation only, we can predict the number of moles of oxygen to react with $3.40mol$ ${C_3}{H_7}OH$. Here actually we proceed with calculation by analyzing the proportion in which reactants interact to form products which is in relation with stoichiometry. We have to use the mole ratio of components which were asked to compare and we will use them as conversion factors.

Complete step-by-step answer:
First of all we have to know whether the equation is balanced before going to the calculation. Let us check this at first. The given equation is as follows,
$2{C_3}{H_7}OH + 9{O_2} \to C{O_2} + 8{H_2}O$
Here we can see it is not balanced as there are six carbon on the left side but only one carbon on right side. Let us balance them properly first.
Let us look on to left side of equation which is as follows,
$C = 6$
$H = 14$
$O = 2 + 18$
Next we have to look on to right side of equation which is as follows,
$C = 1$
$H = 16$
$O = 2 + 8$

Since it is not balanced, we have to balance it using any balancing method as the oxidation method is more preferable to use. On balancing, we will be having the equation as follows,
${C_3}{H_7}OH + \dfrac{9}{2}{O_2} \to 3C{O_2} + 4{H_2}O$
Let us check on left side which is as follows,
$C = 3$
$H = 8$
$O = 1 + (2 \times \dfrac{9}{2}) = 10$
Next let us check on right side which is as follows,
$C = 1 \times 3 = 3$
$H = 2 \times 4 = 8$
$O = (2 \times 3) + (1 \times 4) = 10$

Since the reactant side is equal to the product side, the equation is balanced. Now, let us proceed the calculation to get the answer.

Always remember to balance the equation before going to calculation as it is mainly depending on the correct coefficient in the given chemical equation.
Here our aim is to find the number of moles of ${O_2}$ which is needed to react with $3.40moles$ of ${C_3}{H_7}OH$.

From the balanced equation above we can understand that, for one mole of ${C_3}{H_7}OH$ , we need $\dfrac{9}{2}moles$ of ${O_2}$ to react with.
Therefore, for $3.40moles$ of ${C_3}{H_7}OH$, it will be as follows,
$3.40moles{C_3}{H_7}OH \times \dfrac{{4.5mol{O_2}}}{{1mol{C_3}{H_7}OH}}$
$= 15.3mol{O_2}$

Hence, $15.3mole$ of oxygen are required to react with $3.40moles$ of ${C_3}{H_7}OH$.

Note: First of all we have to keep in mind that, while dealing with this type of question we have to remember to always balance at first before proceeding to calculation. This is because we will be having a tendency to compute directly by recognizing the coefficient from the given skeletal equation itself , which will be wrong for sure.

We have to note that actually the correct calculation depends on the correct coefficients in the chemical equation and hence we should be aware of balancing it at the first moment itself. It is actually indirectly depending on conservation law of mass as chemical reactions are those that just happen by rearrangement of atoms that can change physical and chemical properties of compounds involved.