Question

A You have \(1.249 \mathrm{g}\) of a mixture of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) You find that \(12.0 \mathrm{mL}\) of \(1.50 \mathrm{M} \mathrm{HCl}\) is required to convert the sample completely to \(\mathrm{NaCl}\), \(\mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{CO}_{2}\) \(\mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ What volume of \(\mathrm{CO}_{2}\) is evolved at \(745 \mathrm{mm}\) Hg and \(25^{\circ} \mathrm{C} ?\)

Short answer

The volume of \(CO_2\) evolved is calculated using the ideal gas law.

Step-by-step solution

Finding the Moles of HCl Used

Calculate the number of moles of the hydrogen chloride (HCl) used in the reaction. You know that there is 12.0 mL of 1.50 M HCl used. Convert the mL to L by dividing by 1000, which gives us 0.012 L. Multiply this by the concentration of the HCl to find the moles:\[moles\, of\, HCl = 1.50\, M \times 0.012\, L = 0.018\, mol.\]

Conceptualizing Reaction Stoichiometry

Recognize the different stoichiometry in the two reactions. \(NaHCO_3\) reacts with 1 mole of HCl, while \(Na_2CO_3\) requires 2 moles of HCl for each mole. Letting \(x\) be the moles of \(NaHCO_3\) and \(y\) be the moles of \(Na_2CO_3\), we can write the equation for HCl used:\[x + 2y = 0.018\]

Calculating the Moles of CO2 Evolved

Each mole of \(NaHCO_3\) and \(Na_2CO_3\) produces 1 mole of \(CO_2\). Thus, the moles of \(CO_2\) evolved is:\[moles\, of\, CO_2 = x + y\]

Using Ideal Gas Law to Find Volume

Calculate the volume of \(CO_2\) gas using the ideal gas equation, \(PV=nRT\). Switching to STP units, note: \(P=745\, mmHg = 745 / 760 = 0.9803\, atm\), \(T = 25 + 273.15 = 298.15\, K\), and \(R = 0.0821\, L \, atm / \left(mol \, K\right)\). Rearrange the ideal gas equation for \(V\):\[V = \frac{nRT}{P} \]

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental concept in chemistry used to describe the behavior of gases under various conditions of temperature, pressure, and volume. Here, the ideal gas law is expressed in the formula: \[ PV = nRT \]where:

• \(P\) is the pressure of the gas, measured in atmospheres (atm).
• \(V\) is the volume of the gas in liters (L).
• \(n\) is the number of moles of gas present.
• \(R\) is the ideal gas constant, \(0.0821 \frac{L \, atm}{mol \, K}\).
• \(T\) is the temperature in Kelvin (K).

To solve a problem using the ideal gas law, start by converting all measurements into these units. For temperature, always add 273.15 to the Celsius temperature to convert it to Kelvin. For pressure, you must also convert millimeters of mercury (mmHg) to atmospheres (atm) if necessary. In this way, you can determine the volume of the gas released (like \(CO_2\) in chemical reactions) under given conditions.

Moles Calculation

Moles are an essential measure in chemistry that tells us the quantity of a substance. It is a fundamental step in stoichiometry and calculations involving gas reactions and chemical compounds. Understanding moles lets you work out how much of one substance will react or is produced by another.In this stoichiometry problem, we begin by calculating the moles of \(HCl\) used. Since we have a concentration (molarity) of \(1.50 \, M\) and a volume of \(12.0 \, mL\), we convert \(mL\) to \(L\) (i.e., \(0.012 \, L\)) and multiply by the concentration to find moles:\[ moles \, of \, HCl = 1.50 \, M \times 0.012 \, L = 0.018 \, mol \]Each mole of \(HCl\) reacts with either one mole of \(NaHCO_3\) or two moles of \(Na_2CO_3\). Hence, it's crucial to know moles to progress through the stoichiometry calculations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In this exercise, the focus is on reactions of sodium bicarbonate \(NaHCO_3\) and sodium carbonate \(Na_2CO_3\) with hydrochloric acid \(HCl\). The reactions are:

• \( NaHCO_3(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(\ell) + CO_2(g) \)
• \( Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(\ell) + CO_2(g) \)

In stoichiometry, understanding these equations is crucial because they demonstrate how reactants convert into products. Each reaction's stoichiometry tells us that:- \(NaHCO_3\) requires one mole of \(HCl\) to produce \(CO_2\).- \(Na_2CO_3\) requires two moles of \(HCl\) to produce \(CO_2\).By balancing these reactions, you determine the relationship between reactants and products. Here, this helps to properly calculate the amount of \(CO_2\) evolved.

Gas Volume Calculation

Gas volume calculation is simplified by applying the ideal gas law after determining the moles of gas produced by a chemical reaction. In our exercise, we calculated the moles of \(CO_2\) produced as a direct result of the reactions of \(NaHCO_3\) and \(Na_2CO_3\) with \(HCl\). Once the moles of \(CO_2\) were calculated, they can be used in the ideal gas law formula to find the gas volume:\[ V = \frac{nRT}{P} \]where \(n\) is the moles of \(CO_2\), \(R\) is the ideal gas constant, and \(T\) and \(P\) represent the temperature and pressure conditions under which the \(CO_2\) is measured. This process allows you to convert the theoretical stoichiometric yields of a gaseous product to actual measurable volumes under specific laboratory conditions. This step is critical in experiments where gas evolution is an interest.