Question

Consider the following unbalanced chemical equation: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ What volume of oxygen gas, measured at \(28^{\circ} \mathrm{C}\) and 0.976 atm, is needed to react with \(5.00 \mathrm{g}\) of the the perents \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?\) What volume of each product is produced under the same conditions?

Short answer

Under the given conditions, 4.30 L of oxygen gas is needed to react with 5.00 g of glucose, producing 4.30 L of carbon dioxide and 4.30 L of water vapor.

Step-by-step solution

Balance the reaction equation

Balancing the oxidation numbers of each element in the reaction equation, we obtain: $$ \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}(s)+6\mathrm{O}_{2}(g)\rightarrow 6\mathrm{CO}_{2}(g)+6\mathrm{H}_{2}\mathrm{O}(g) $$

Convert the given mass of glucose to moles

Given mass of glucose = 5.00 g Molecular weight of glucose (\(C_6H_{12}O_6\)) = (6×12.01 g/mol) + (12×1.01 g/mol) + (6×16.00 g/mol) = 180.18 g/mol Moles of glucose = \(\frac{5.00\,\text{g}}{180.18\,\text{g/mol}}=0.0277\,\text{mol}\)

Determine the stoichiometric ratios

Now let's analyze the balanced equation to find the stoichiometric ratios: 1 molecule of glucose reacts with 6 molecules of oxygen to produce 6 molecules of carbon dioxide and 6 molecules of water. Therefore, mole ratios are: $$ 1\,\text{mol}\,\mathrm{C_6H_{12}O_6} : 6\,\text{mol}\,\mathrm{O_2} : 6\,\text{mol}\,\mathrm{CO_2} : 6\,\text{mol}\,\mathrm{H_2O} $$

Calculate the moles of oxygen required and the moles of products produced

Using the stoichiometric ratios, we can calculate: Moles of \(O_2\) required = 0.0277 mol (glucose) × \(\frac{6\,\text{mol}\,\mathrm{O_2}}{1\,\text{mol}\,\mathrm{C_6H_{12}O_6}} = 0.166\,\text{mol}\) Moles of \(CO_2\) produced = 0.0277 mol (glucose) × \(\frac{6\,\text{mol}\,\mathrm{CO_2}}{1\,\text{mol}\, \mathrm{C_6H_{12}O_6}} = 0.166\,\text{mol}\) Moles of \(H_2O\) produced = 0.0277 mol (glucose) × \(\frac{6\,\text{mol}\, \mathrm{H_2O}}{1\,\text{mol}\,\mathrm{C_6H_{12}O_6}} = 0.166\,\text{mol}\)

Use the Ideal Gas Law to obtain the volume of substances

The ideal gas law states that \(PV = nRT\), where P = pressure, V = volume, n = number of moles, R = ideal gas constant, and T = temperature. Given: Temperature = \((28+273.15) \,\text{K} = 301.15\,\text{K}\) Pressure = 0.976 atm Ideal gas constant (R) = 0.0821 L atm/mol K Use these values to calculate the volume of the oxygen required and the volumes of each product produced: Volume of \(O_2\) required = \(\frac{(0.166\,\text{mol}) (0.0821\,\text{L atm/mol K}) \cdot (301.15\,\text{K})}{0.976\,\text{atm}} = 4.30\,\text{L}\) Volume of \(CO_2\) produced = \(\frac{(0.166\,\text{mol}) (0.0821\,\text{L atm/mol K}) \cdot (301.15\,\text{K})}{0.976\,\text{atm}} = 4.30\,\text{L}\) Volume of \(H_2O\) produced = \(\frac{(0.166\,\text{mol}) (0.0821\,\text{L atm/mol K}) \cdot (301.15\,\text{K})}{0.976\,\text{atm}} = 4.30\,\text{L}\) So, under the given conditions, 4.30 L of oxygen gas is needed to react with 5.00 g of glucose, producing 4.30 L of carbon dioxide and 4.30 L of water vapor.

Key concepts

Balancing Chemical Equations

The first step in understanding chemical stoichiometry is to learn about balancing chemical equations. This process involves adjusting the coefficients of the reactants and products so that the number of atoms for each element is the same on both sides of the reaction equation. This law is rooted in the conservation of mass, which states that matter cannot be created or destroyed.

To balance equations, start by counting the number of atoms of each element on both sides of the reaction. If they are unequal, multiply the molecules of the substances by appropriate coefficients to achieve balance. For instance, in the reaction equation given in the exercise \( \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \) it was initially unbalanced. Upon examination, it's clear that 6 oxygen molecules (\( \mathrm{O}_2 \) are required to balance the equation, which then looks like \( \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}(s)+6\mathrm{O}_{2}(g)\rightarrow 6\mathrm{CO}_{2}(g)+6\mathrm{H}_{2}\mathrm{O}(g) \).

Mole Concept

The mole concept is a cornerstone in stoichiometry, helping chemists relate quantities of substances. A mole is defined as the amount of a substance that contains exactly \(6.022 \times 10^{23}\) particles, which could be atoms, molecules, ions, or electrons. This is known as Avogadro's number.

When dealing with chemical reactions, the mole allows us to convert between mass and number of molecules easily because of its tie to the molecular weight of a substance. The molecular weight, typically in grams per mole (\(g/mol\)), shows the weight of one mole of a substance. For example, in our original exercise's step 2, the molecular weight of glucose (\(C_6H_{12}O_6\)) is calculated to be 180.18 \(g/mol\), which then allows us to convert 5.00 grams of glucose into moles for further calculations in the problem.

Ideal Gas Law

Finally, the ideal gas law is an equation of state for a hypothetical ideal gas which is a good approximation to many gases under many conditions. The law is represented as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

In the context of our exercise, after calculating the moles of oxygen, carbon dioxide, and water vapor produced, we apply the ideal gas law to each to determine their volumes under the given conditions of temperature and pressure. This was the final step in our solution, where we used the known values of \(T = 301.15 K\), \(P = 0.976 atm\), and \(R = 0.0821 L\ atm/mol K\) to find the volumes of \(O_2\), \(CO_2\), and \(H_2O\), which all appeared to be 4.30 L.

Understanding the ideal gas law is essential for predicting how gases will behave under different conditions and is a fundamental concept in chemistry.