Question 814cd

May 25, 2016

The reaction will produce $24$ moles of ${\text{CO}}_{2}$.

Explanation:

Start by taking a look at the chemical equation given to you

${\text{C"_ 6"H"_ 12"O"_ (6(aq)) + 6"O"_ (2(g)) rightleftharpoons color(red)(6)"CO"_ (2(g)) + 6"H"_ 2"O}}_{\left(l\right)}$

In order to show that this chemical equation is balanced, you must make sure that all the atoms that are present on the reactants' side are also present on the products' side.

Now, chemical equations make use of subscripts to denote number of atoms in a molecule and coefficients to denote number of molecules.

For example, you have

$\text{O"_color(blue)(2) = color(blue)(2) xx "O}$

Here each molecule of ${\text{O}}_{2}$ contains $2$ atoms of oxygen

$6 \text{O"_2 = 6 xx "O"_color(blue)(2) = 6 xx [color(blue)(2) xx "O"] = 12 xx "O}$

Here $6$ molecules will contain a total of $12$ atoms of oxygen.

With this in mind, inspect both sides of the equation and make an inventory of all the atoms that appear.

The reactants' side contains

• six atoms of carbon, $6 \times \text{C}$
• twelve atoms of hydrogen, $12 \times \text{H}$
• eighteen atoms of oxygen, $6 \times \text{O" + 6 xx [2 xx "O"] = 18 xx "O}$

The products' side contains

• six atoms of carbon, $6 \times \text{C}$
• twelve atoms of hydrogen, 6 xx [2xx"H"] = 12 xx "H"
• eighteen atoms of oxygen, 6 xx [2xx"O"] + 6 xx"O" = 18 xx "O"#

Since both sides of the equation contain the same number of atoms of each element, you can conclude that the chemical equation given to you is balanced.

Now, you know that you have $4$ moles of glucose. Since oxygen gas is said to be in excess, you don't to worry about the number of moles of ${\text{O}}_{2}$.

In other words, the reaction will consume all the moles of glucose given to you, i.e. $4$ moles.

Notice that you have a $1 : \textcolor{red}{6}$ mole ratio between glucose and carbon dioxide, ${\text{CO}}_{2}$. This means that the reaction produces $\textcolor{red}{6}$ moles of carbon dioxide for every moles of glucose that reacts.

In your case, you will have

$4 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles C"_6"H"_12"O"_6))) * (color(red)(6)color(white)(a)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_6"H"_12"O"_6)))) = color(green)(|bar(ul(color(white)(a/a)"24 moles CO}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$